The angle of depression of a car parked on the road from the top of a $75$ m high tower is $30^{\circ}$. The distance of the car from the base of the tower is :
From a point on the ground, which is $30 \text{ m}$ away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be $60^\circ$. The height (in metres) of the tower is:
From a point on the ground, which is $30 \text{ m}$ away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be $60^\circ$. The height (in metres) of the tower is :
A ladder $14$ m long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, then the height of the wall is :
A kite is flying at a height of $150$ m from the ground. It is attached to a string inclined at an angle of $30^\circ$ to the horizontal. The length of the string is:
Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is textbf{not} the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A): A ladder leaning against a wall, stands at a horizontal distance of $6 \operatorname{m}$ from the wall. If the height of the wall up to which the ladder reaches is $8 \operatorname{m}$, then the length of the ladder is $10 \operatorname{m}$. Reason (R): The ladder makes an angle of $60^{\circ}$ with the ground.
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Sol. (C) Assertion (A) is true, but Reason (R) is false.
A $30$ m long rope is tightly stretched and tied from the top of pole to the ground. If the rope makes an angle of $60^\circ$ with the ground, the height of the pole is :
A $30$ m long rope is tightly stretched and tied from the top of pole to the ground. If the rope makes an angle of $60^{\circ}$ with the ground, the height of the pole is :
An observer 1.8 m tall stands away from a chimney at a distance of 38.2 m along the ground. The angle of elevation of top of chimney from the eyes of observer is $45^\circ$. The height of chimney above the ground is
A peacock sitting on the top of a tree of height 10 m observes a snake moving on the ground. If the snake is $10\sqrt{3}$ m away from the base of the tree, then angle of depression of the snake from the eye of the peacock is
The length of the shadow of a tower on the plane ground is $\sqrt{3}$ times the height of the tower. Find the angle of elevation of the sun.
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Let AB be the tower of height 'h'. $\therefore$ AC = $\sqrt{3}$ h In $\triangle ABC$, $\tan \theta = \frac{AB}{AC} = \frac{h}{\sqrt{3} h}$ $\Rightarrow \tan \theta = \frac{1}{\sqrt{3}}$ $\Rightarrow \theta = 30^\circ$
The angle of elevation of the top of a tower from a point on the ground which is $30$ m away from the foot of the tower, is $30^\circ$. Find the height of the tower.
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Height of tower = AB In $\triangle ABC$, $\tan 30^\circ = \frac{AB}{30}$ AB = $\frac{30}{\sqrt{3}} = 10\sqrt{3}$ $\therefore$ Height of Tower is $10 \sqrt{3}$ m
Find the length of the shadow on the ground of a pole of height $18$ m when angle of elevation $\theta$ of the sun is such that $\tan \theta = \frac{6}{7}$.
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Pole of height $AB = 18$ m $AP = \text{length of shadow}$ In $\triangle APB$, $\tan \theta = \frac{18}{AP}$ $\frac{6}{7} = \frac{18}{AP}$ $\Rightarrow AP = 21$ m
It is found that on walking $20$ m towards a chimney in a horizontal line through its base, the elevation of its top changes from $30^\circ$ to $60^\circ$. The height of the chimney is :
The angle of elevation of the top of a tower 24 m high from the foot of another tower in the same plane is $60^\circ$. The angle of elevation of the top of second tower from the foot of the first tower is $30^\circ$. Find the distance between two towers and the height of the other tower. Also, find the length of the wire attached to the tops of both the towers.
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1 mark for correct figure Let $AB$ and $CD$ be the given towers. $\tan 30^\circ = \frac{h}{x} \Rightarrow x = h\sqrt{3}$ (i) $\tan 60^\circ = \sqrt{3} = \frac{24}{x} \Rightarrow x = \frac{24}{\sqrt{3}}$ or $8\sqrt{3}$ (ii) using (i) and (ii) $x=8\sqrt{3}$ and $h=8$ length of wire = $\sqrt{BE^2 + X^2} = \sqrt{256 + 192} = \sqrt{448}$ m = $8\sqrt{7}$ m
Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two Sections A and B. Tower is supported by wires from a point O. Distance between the base of the tower and point O is $36$ cm. From point O, the angle of elevation of the top of the Section B is $30^{\circ}$ and the angle of elevation of the top of Section A is $45^{\circ}$. Based on the above information, answer the following questions : (i) Find the length of the wire from the point O to the top of Section B. (ii) Find the distance AB. OR Find the area of $\triangle OPB$. (iii) Find the height of the Section A from the base of the tower.
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(i) In $\triangle OBP$, $\cos 30^{\circ} = \frac{OP}{OB}$ $\frac{\sqrt{3}}{2} = \frac{36}{OB} \Rightarrow OB = \frac{72}{\sqrt{3}} = 24\sqrt{3}$ cm (ii)In $\triangle OBP$, $\tan 30^{\circ} = \frac{PB}{36} \Rightarrow PB = \frac{36}{\sqrt{3}} = 12\sqrt{3}$ In $\triangle OAP$, $\tan 45^{\circ} = \frac{AP}{36} \Rightarrow AP = 36$ cm $AB = AP - PB = 36 - 12\sqrt{3} = 12(3 - \sqrt{3})$ cm OR (ii)Area of $\triangle OPB = \frac{1}{2} \times OP \times PB$ $= \frac{1}{2} \times 36 \times 12\sqrt{3} = 216\sqrt{3}$ cm$^2$ (iii) $AP = 36$ cm
Due to short circuit, a fire has broken out in New Home Complex. Two buildings, namely X and Y have mainly been affected. The fire engine has arrived and it has been stationed at a point which is in between the two buildings. A ladder at point O is fixed in front of the fire engine. The ladder inclined at an angle $60^\circ$ to the horizontal is leaning against the wall of the terrace (top) of the building Y. The foot of the ladder is kept fixed and after some time it is made to lean against the terrace (top) of the opposite building X at an angle of $45^\circ$ with the ground. Both the buildings along with the foot of the ladder, fixed at 'O' are in a straight line. Based on the above given information, answer the following questions : (i) Find the length of the ladder. (ii) Find the distance of the building Y from point 'O', i.e. OA. (iii) (a) Find the horizontal distance between the two buildings. OR (b) Find the height of the building X.
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(i) In $\triangle OAP$, $\frac{OP}{12\sqrt{3}} = \cosec 60^\circ = \frac{2}{\sqrt{3}}$ $\Rightarrow OP = 24$ m $\therefore$ Length of ladder is $24$ m (ii) In $\triangle OAP$, $\frac{OA}{12\sqrt{3}} = \cot 60^\circ = \frac{1}{\sqrt{3}}$ $\Rightarrow OA = 12$ m $\therefore$ the distance of the building Y from point O ie.,OA is $12$ m (iii) (a) OP = OR = $24$ m $\therefore$ In $\triangle OCR$, $\frac{OC}{24} = \cos 45^\circ = \frac{1}{\sqrt{2}}$ $\Rightarrow OC = 12\sqrt{2}$ m $\therefore$ distance between two buildings $=$ OA + OC $= (12 + 12\sqrt{2})$ m or $12(1 + \sqrt{2})$ m OR (iii) (b) OP = OR = $24$ m $\therefore$ In $\triangle OCR$, $\frac{RC}{24} = \sin 45^\circ = \frac{1}{\sqrt{2}}$ $\Rightarrow RC = 12\sqrt{2}$ m $\therefore$ height of building X is $12\sqrt{2}$ m
Case Study – 2 A class VI student went to a park and went up the slide to play. The angle of elevation of the slide is $30^{\circ}$, but the base from which the angle of elevation is measured is $50$ cm above the ground level and the distance of this point from the bottom of the staircase (which is vertical) is $4\sqrt{3}$ m. Based on the above information, answer the following questions: (i) Write the angle of depression from the top of the slide to its base. (ii) (a) Find the height of the staircase. OR (b) Find the length of the slide. (iii) Will the angle of elevation increase or decrease if the staircase was made taller?
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Sol. Let AB be the staircase. (i) $30^{\circ}$ (ii) (a) $\text{tan } 30^{\circ} = \frac{h-0.5}{4\sqrt{3}}$ $\Rightarrow h = 4.5$ So, height of the staircase is $4.5$ m OR (b) $\text{cos } 30^{\circ} = \frac{4\sqrt{3}}{l} = \frac{\sqrt{3}}{2}$ $\Rightarrow l = 8$ So, length of the slide is $8$ m. (iii) Angle of elevation will increase.
Case Study – 2 Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two stations A and B (B vertically below A). The tower is supported by wires AO and BO from a point O on the ground. Distance between the base C of the tower and the point O is $36$ m. From O, the angles of elevation of the tops of station B and station A are $30^\circ$ and $45^\circ$ respectively. Based on the above, answer the following questions : (i) Find the height of station B. (ii) Find the height of station A. (iii) (a) Find the length of the wire OA. OR (b) Find the length of the wire OB.
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(i) $$\begin{aligned}& \tan 30^\circ = \frac{BC}{36} = \frac{1}{\sqrt{3}} \\ & \Rightarrow BC = 12 \sqrt{3}\end{aligned}$$ m (ii) $$\begin{aligned}& \tan 45^\circ = \frac{AC}{36} = 1 \\ & \Rightarrow AC = 36\end{aligned}$$ m (iii) (a) $$\begin{aligned}& \sec 45^\circ = \frac{OA}{36} = \sqrt{2} \\ & \Rightarrow OA = 36\sqrt{2}\end{aligned}$$ m OR (iii) (b) $$\begin{aligned}& \sec 30^\circ = \frac{OB}{36} = \frac{2}{\sqrt{3}} \\ & \Rightarrow OB = 24\sqrt{3}\end{aligned}$$ m
SECTION E This section has $3$ case study based questions carrying $4$ marks each. Case Study -1 The International Kite Festival takes place every year on $14^{th}$ January. The main attractions of the festival include national and international Kite Flyers' Parade, kite flying, traditional stalls etc. On this day, few kite flyers, had assembled at a point 'O' on the ground. The position of $3$ kites A, B, C was such that A and B were at the same vertical height of $40$ m from the ground level. The angles of elevation of A, B and C from O were $60^{\circ}, 45^{\circ}$ and $30^{\circ}$ respectively. A vertical tower, SD has been erected at point S and a camera is set at the top of the tower for photography. Based on the information given above, answer the following questions : (i) What is the length of the string of the kite at A? (ii) If the length of the string of kite at C is $40$ m, then find the height of that kite C from the ground. (iii) (a) What is the horizontal distance between the kites at A and B? OR (iii) (b) If the angle of depression of the kite at A is $30^{\circ}$ from the camera at D and the distance between A and D is $60$ m, then find the height of the tower.
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(i) $\sin 60^{\circ} = \frac{40}{OA} = \frac{\sqrt{3}}{2}$ $\Rightarrow OA = \frac{80}{\sqrt{3}}$ m or $\frac{80\sqrt{3}}{3}$ m (ii) $\sin 30^{\circ} = \frac{RC}{40} = \frac{1}{2}$ $\Rightarrow RC = 20$ m (iii) (a) $\tan 45^{\circ} = \frac{40}{OQ} = 1$ $\Rightarrow OQ = 40$ m Also, $\tan 60^{\circ} = \frac{40}{OP} = \sqrt{3}$ $\Rightarrow OP = \frac{40}{\sqrt{3}}$ m or $\frac{40\sqrt{3}}{3}$ m AB = PQ = $\left(40 + \frac{40}{\sqrt{3}}\right)$ m or $\left(40 + \frac{40\sqrt{3}}{3}\right)$ m OR (b) $\sin 30^{\circ} = \frac{h}{60} = \frac{1}{2}$ $\Rightarrow h = 30$ m Height of the tower = $40 + 30 = 70$ m
Case Study - 3: Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be $60^\circ$. Then, she climbed a nearby observation deck, $40$ metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be $45^\circ$. (i) If $CD$ is $h$ metres, find the distance $BD$ in terms of '$h$'. (ii) Find distance $BC$ in terms of '$h$'. (iii) (a) Find the height $CE$ of the lighthouse [Use $\sqrt{3} = 1.73$]. OR (iii) (b) Find distance $AE$, if $AC = 100$ m.
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(i) $\frac{h}{BD} = \tan 45^\circ = 1 \implies BD = h$ m. (ii) $\frac{h}{BC} = \sin 45^\circ = \frac{1}{\sqrt{2}} \implies BC = \sqrt{2}h$ m. (iii)(a) $\tan 60^\circ = \frac{EC}{AE} \implies \sqrt{3} = \frac{h + 40}{h} \implies h = 54.6$ m. $CE = 94.6$ m. (iii)(b) $\cos 60^\circ = \frac{AE}{AC} \implies \frac{1}{2} = \frac{AE}{100} \implies AE = 50$ m.
Passenger boarding stairs, sometimes referred to as boarding ramps, stair cars or aircraft steps, provide a mobile means to travel between the aircraft doors and the ground. Larger aircraft have door sills 5 to 20 feet (1 foot = 30 cm) high. Stairs facilitate safe boarding and de-boarding. An aircraft has a door sill at a height of 15 feet above the ground. A stair car is placed at a horizontal distance of 15 feet from the plane. Based on given information, answer the questions given in part (i) and (ii). (i) Find the angle at which stairs are inclined to reach the door sill 15 feet high above the ground. (ii) Find the length of stairs used to reach the door sill. Further, answer any one of the following questions: (iii) (a) If the 20 feet long stairs is inclined at an angle of $60^\circ$ to reach the door sill, then find the height of the door sill above the ground. (use $\sqrt{3} = 1.732$) OR (iii) (b) What should be the shortest possible length of stairs to reach the door sill of the plane 20 feet above the ground, if the angle of elevation cannot exceed $30^\circ$? Also, find the horizontal distance of base of stair car from the plane.
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(i) $\tan \theta = \frac{15}{15} = 1 \implies \theta = 45^\circ$ (ii) $\frac{15}{l} = \sin 45^\circ \implies l = 15\sqrt{2}$ ft. or 21.21 ft. approx. (iii) (a) $\frac{h}{20} = \sin 60^\circ = \frac{\sqrt{3}}{2} \implies h = 10\sqrt{3} = 17.32$ ft. (iii) (b) $\frac{20}{l} = \sin 30^\circ = \frac{1}{2} \implies l = 40$ ft. $\frac{20}{x} = \tan 30^\circ = \frac{1}{\sqrt{3}} \implies x = 20\sqrt{3}$ ft. or 34.64 ft. approx.
The Statue of Unity situated in Gujarat is the world's largest Statue which stands over a 58 m high base. As part of the project, a student constructed an inclinometer and wishes to find the height of Statue of Unity using it. He noted following observations from two places: Situation - I: The angle of elevation of the top of Statue from Place A which is $80\sqrt{3}$ m away from the base of the Statue is found to be $60^\circ$. Situation - II: The angle of elevation of the top of Statue from a Place B which is 40 m above the ground is found to be $30^\circ$ and entire height of the Statue including the base is found to be 240 m. Based on given information, answer the following questions: (i) Represent the Situation - I with the help of a diagram. (ii) Represent the Situation - II with the help of a diagram. (iii) (a) Calculate the height of Statue excluding the base and also find the height including the base with the help of Situation - I. OR (iii) (b) Find the horizontal distance of point B (Situation - II) from the Statue and the value of $\tan \alpha$, where $\alpha$ is the angle of elevation of top of base of the Statue from point B.
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(i) Correct figure (1 mark). (ii) Correct figure (1 mark). (iii) (a) In $\Delta ACQ$, $\frac{QC}{AC} = \tan 60^\circ = \sqrt{3} \Rightarrow QC = 240$ m. Height of statue including base = 240 m. Height of statue excluding base = $240 - 58 = 182$ m (1 + 1 marks). OR (iii) (b) $QR = 240 - 40 = 200$ m. In $\Delta QRB$, $\frac{QR}{RB} = \tan 30^\circ = \frac{1}{\sqrt{3}}$. Horizontal distance $RB = 200\sqrt{3}$ m ($\frac{1}{2} + \frac{1}{2}$ marks). Correct figure ($\frac{1}{2}$ mark). In $\Delta PRB$, $\tan \alpha = \frac{PR}{BR} = \frac{18}{200\sqrt{3}}$ or $\frac{3\sqrt{3}}{100}$ ($\frac{1}{2}$ mark).
Clinometer is a tool that is used to measure the angle of elevation. We can use the clinometer to measure the height of tall things that you can't possibly reach. With the help of a clinometer, Harish measured the angle of elevation of the roof of a building from a point $P$ on the ground as $45^\circ$. On the same wall, at some height below the top, there was a society logo, whose angle of elevation from the same point $P$ was measured as $30^\circ$. The point $P$ is at a distance of $24$ m from the base of the building. Based on the above information, answer the following questions : (a) (i) What is the height of the building logo from the ground ? OR (ii) What is the height of the building from the ground ? (b) What is the aerial (slant) distance of point $P$ from the top of the building ? (c) If $\theta$ is the angle of elevation of the top of building when the point $P$ is moved $9$ m towards the base of the building, then, find $\tan \theta$.
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(a) (i) $\frac{h}{24} = \tan 30^\circ = \frac{1}{\sqrt{3}} \implies h = \frac{24}{\sqrt{3}}$ or $8\sqrt{3}$ m OR (ii) $\frac{H}{24} = \tan 45^\circ = 1 \implies H = 24$ m (b) Slant distance = $AP = \sqrt{24^2 + 24^2} = 24\sqrt{2}$ m (c) $CP = (24 - 9) = 15$ m. $\tan \theta = \frac{24}{15}$ or $\frac{8}{5}$
As observed from the top of a lighthouse, $100$ m above sea level, the angle of depression of a ship, sailing directly towards it, changes from $30^\circ$ to $45^\circ$. Determine the distance travelled by the ship during the period of observation. (Use $\sqrt{3}= 1.732$)
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Correct figure. In $\triangle ABC$ $\frac{100}{BC} = \tan 45^\circ = 1$ $\Rightarrow BC = 100$ In $\triangle ABD$ $\frac{100}{BD} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ $\Rightarrow BD = 100 \sqrt{3}$ $\Rightarrow 100 + CD = 100 \sqrt{3}$ $\Rightarrow CD = 100 \sqrt{3} - 100 = 100 (1.732 - 1) = 73.2$ Hence, distance travelled by the ship during the period of observation is $73.2$ m
From the top of a $60$ m high building, the angles of depression of the top and bottom of a cable tower are observed to be $45^\circ$ and $60^\circ$ respectively. Find the height of the tower. (Use $\sqrt{3} = 1.73$)
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Sol. Correct figure (2 Marks) Let height of the tower be '$h$' m and ED = BC = '$x$' m In $\triangle AED$ $\frac{60-h}{x} = \tan 45^\circ = 1$ $\Rightarrow 60-h=x$ -----(1) (1/2 Mark) In $\triangle ABC$ $\frac{60}{x} = \tan 60^\circ = \sqrt{3}$ $\Rightarrow 60 = \sqrt{3} x$ (1/2 Mark) $\Rightarrow 60 = \sqrt{3} (60 - h)$ (1 Mark) $\Rightarrow h = 60 - \frac{60}{\sqrt{3}} = 60 - 20\sqrt{3} = 20(3-\sqrt{3})$ (1/2 Mark) $\Rightarrow h = 20 (3-1.73)$ $\Rightarrow h = 25.4$ (1/2 Mark) Hence, height of the tower is $25.4$ m.
A spherical balloon of radius $r$ subtends an angle of $60^\circ$ at the eye of an observer. If the angle of elevation of its centre is $45^\circ$ from the same point, then prove that height of the centre of the balloon is $\sqrt{2}$ times its radius.
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1 mark for correct figure Let Point B represents observer. $\therefore \angle QBP = 60^\circ$; $\angle ABO = 45^\circ$ Using geometry $\angle PBO = \frac{1}{2} \times 60^\circ = 30^\circ$ Now, $\frac{r}{OB} = \sin 30^\circ = \frac{1}{2} \Rightarrow OB = 2r$ (i) Also $\frac{OA}{OB} = \sin 45^\circ = \frac{1}{\sqrt{2}} \Rightarrow OB = OA \sqrt{2}$ (ii) Using (i) and (ii) $OA = \sqrt{2} r$ or height of center of balloon = $\sqrt{2} r$ units
A ladder set against a wall at an angle $45^{\circ}$ to the ground. If the foot of the ladder is pulled away from the wall through a distance of 4 m, its top slides a distance of 3 m down the wall making an angle $30^{\circ}$ with the ground. Find the final height of the top of the ladder from the ground and length of the ladder.
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1 for correct figure $\sin 45^{\circ} = \frac{AB}{BD} = \frac{h+3}{BD}$ $\Rightarrow BD = (h + 3) \sqrt{2}$ ----- (i) $\sin 30^{\circ} = \frac{h}{CE} = \frac{1}{2}$ $\Rightarrow CE = 2h$ ----- (ii) length of ladder remains same Therefore $BD = CE \Rightarrow (h + 3) \sqrt{2} = 2h$ $\Rightarrow h = \frac{3\sqrt{2}}{2-\sqrt{2}} = 3(\sqrt{2} + 1)$ Final height of the top of the ladder = $3(\sqrt{2} + 1) \text{ m}$ and length of ladder = $2h = 6(\sqrt{2} + 1) \text{ m}$
An aeroplane when flying at a height of $3000$ m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are $60^{\circ}$ and $45^{\circ}$ respectively. Find the vertical distance between the aeroplanes at that instant. Also, find the distance of first plane from the point of observation. (Take $\sqrt{3} = 1.73$)
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1 mark for figure Let planes be located at points $C$ and $$\begin{aligned}& D \\ & \tan 45^{\circ} = 1 = \frac{3000 - h}{x} \Rightarrow x = 3000-h \quad \text{------(i)} \\ & \tan 60^{\circ} = \sqrt{3} = \frac{3000}{x} \Rightarrow x = 1000 \sqrt{3} \quad \text{------(ii)} \\ & \text{Using (i) and (ii) } h = 3000 - 1730 \\ & = 1270 \\ & \therefore \text{vertical distance between the aeroplanes } = 1270 \text{ m} \\ & \text{Also } \sin 60^{\circ} = \frac{\sqrt{3}}{2} = \frac{3000}{z} \Rightarrow z = 2000 \sqrt{3} = 3460 \\ & \text{Distance of the first plane from the point of observation } = 3460 \text{ m}\end{aligned}$$
As observed from the top of a $75$ m high lighthouse from the sea-level, the angles of depression of two ships are $30^\circ$ and $60^\circ$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use $\sqrt{3}= 1.73$)
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1 for correct figure PQ = Height of Light house = $75$ m $\angle XQS = \angle QSP = 30^\circ$ $\angle XQR = \angle QRP = 60^\circ$ R and S are position of ships. In $\triangle PQR$, $\frac{75}{PR} = \tan 60^\circ = \sqrt{3} \Rightarrow PR = \frac{75}{\sqrt{3}} = 25\sqrt{3}$ In $\triangle PQS$, $\frac{75}{PS} = \tan 30^\circ$ PS = $75\sqrt{3}$ $\therefore$ Distance between the ships, RS = PS – PR $= 75 \sqrt{3}-25 \sqrt{3} = 50\sqrt{3}$ $= 50 \times 1.73 = 86.5$ $\therefore$ Distance between the ships is $86.5$ m
From a point on the ground, the angle of elevation of the bottom and top of a transmission tower fixed at the top of $30$ m high building are $30^\circ$ and $60^\circ$, respectively. Find the height of the transmission tower. (Use $\sqrt{3} = 1.73$)
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1 for correct figure Height of building AB = $30$ m BP = transmission tower = h(say) $\angle ACB = 30^\circ$, $\angle ACP = 60^\circ$ In $\triangle ABC$, $\tan 30^\circ = \frac{AB}{AC}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{30}{AC} \Rightarrow AC = 30\sqrt{3}$ In $\triangle APC$, $\tan 60^\circ = \frac{AP}{AC}$ $\sqrt{3}= \frac{30+ h}{30\sqrt{3}}$ $\Rightarrow 30\sqrt{3} \times \sqrt{3}= 30 + h$ $\Rightarrow h = 30 (3 - 1)$ $\Rightarrow h = 60$ $\therefore$ Height of transmission tower = $60$ m
The angle of elevation of the top of a tower $30 \text{ m}$ high from the foot of another tower in the same plane is $60^\circ$ and the angle of elevation of the top of the second tower from the foot of the first tower is $30^\circ$. Find the distance between the two towers and also the height of the other tower.
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1 for correct figure $PQ = \text{height of } 1^{st} \text{ tower} = 30 \text{ m}$ $AB = \text{height of } 2^{nd} \text{ tower} = h \text{ (say)}$ $\angle PAQ = 60^\circ$, $\angle APB = 30^\circ$ Let $AP = x$ In $\triangle APQ$, $\tan 60^\circ = \frac{PQ}{AP} = \frac{30}{x}$ $\Rightarrow x = \frac{30}{\tan 60^\circ} = \frac{30}{\sqrt{3}} = 10\sqrt{3}$ $\therefore \text{Distance between two towers} = 10\sqrt{3} \text{ m}$ In $\triangle APB$, $\tan 30^\circ = \frac{AB}{AP} = \frac{h}{x}$ $\frac{1}{\sqrt{3}} = \frac{h}{10\sqrt{3}}$ $\Rightarrow h = 10$ $\therefore \text{Height of } 2^{nd} \text{ tower} = 10 \text{ m}$
From the top of a tower $100 \text{ m}$ high, a man observes two cars on the opposite sides of the tower with angles of depression $30^\circ$ and $45^\circ$ respectively. Find the distance between the two cars. (Use $\sqrt{3} = 1.73$)
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1 for correct figure $AB = \text{Height of tower} = 100 \text{ m}$ $P$ and $Q$ are position of cars $\angle XBP = \angle APB = 30^\circ$ $\angle YBQ = \angle AQB = 45^\circ$ In $\triangle ABQ$, $\tan 45^\circ = \frac{AB}{AQ} \Rightarrow 1 = \frac{100}{x}$ $\Rightarrow x = 100$ In $\triangle ABP$, $\tan 30^\circ = \frac{AB}{AP} = \frac{100}{y}$ $\frac{1}{\sqrt{3}} = \frac{100}{y} \Rightarrow y = 100\sqrt{3}$ $= 100(1.73) = 173$ Distance between cars = $x + y = 100 + 173 = 273$ $\therefore \text{Distance between cars is } 273 \text{ m}$.
A straight highway leads to the foot of a tower. A man standing on the top of the $75$ m high tower observes two cars at angles of depression of $30^\circ$ and $60^\circ$, which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (use $\sqrt{3} = 1.73$)
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Sol. AB = Height of tower = $75$ m P, Q are positions of cars $\angle XBQ = \angle BQA = 30^\circ$ $\angle XBP = \angle BPA = 60^\circ$ In $\Delta APB$, $\tan 60^\circ = \frac{75}{AP} \Rightarrow AP = \frac{75}{\sqrt{3}} = 25\sqrt{3}$ In $\Delta AQB$, $\tan 30^\circ = \frac{75}{AQ} \Rightarrow AQ = 75 \sqrt{3}$ Distance between the cars = $PQ = AQ – AP$ $= 75\sqrt{3}-25\sqrt{3} = 50\sqrt{3}$ $= 50 \times 1.73 = 86.5$ m
From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $30^\circ$. Determine the height of the tower.
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Sol. Let AC be $h$ m, BC = DE = $7$ m, AB = $(h-7)$ m $\angle AEB = 60^\circ$ and $\angle BEC = 30^\circ$ $\therefore \angle ECD = 30^\circ$ Let CD be $x$ m $\frac{DE}{CD} = \frac{7}{x} = \tan 30^\circ \Rightarrow x = 7\sqrt{3}$ $\Rightarrow BE = 7\sqrt{3}$ Again $\frac{AB}{BE} = \tan 60^\circ$ $\Rightarrow \frac{h-7}{7\sqrt{3}} = \sqrt{3}$ $h - 7 = 21$ $h = 28$ $\therefore$ Height of tower = $28$ m
From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $30^\circ$. Determine the height of the tower.
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Let AC be $h$ m, $BC = DE = 7$ m, $AB = (h−7)$ m $\angle AEB = 60^\circ$ and $\angle BEC = 30^\circ$ $\therefore \angle ECD = 30^\circ$ Let CD be $x$ m $\frac{DE}{CD} = \frac{7}{x} = \tan 30^\circ \Rightarrow x = 7\sqrt{3}$ $\Rightarrow BE = 7\sqrt{3}$ Again $\frac{AB}{BE} = \tan 60^\circ$ $\frac{h-7}{7\sqrt{3}} = \sqrt{3}$ $h - 7 = 21$ $h = 28$ $\therefore \text{Height of tower} = 28$ m
From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $30^\circ$. Determine the height of the tower.
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Let AC be $h$ m, BC = DE = $7$ m, AB = $(h-7)$ m $\angle AEB = 60^\circ$ and $$\begin{aligned}& \angle BEC = 30^\circ \\ & therefore \angle ECD = 30^\circ\end{aligned}$$Let CD be $x$ m $$\begin{aligned}& \frac{DE}{CD} = \tan 30^\circ \Rightarrow \frac{7}{x} = \frac{1}{\sqrt{3}} \Rightarrow x = 7\sqrt{3} \\ & \Rightarrow BE = 7\sqrt{3}\end{aligned}$$Again $$\begin{aligned}& \frac{AB}{BE} = \tan 60^\circ \\ & frac{h-7}{7\sqrt{3}} = \sqrt{3} \\ & h-7 = 7\sqrt{3} \times \sqrt{3} = 7 \times 3 = 21 \\ & h = 28 \\ & therefore\end{aligned}$$ Height of tower = $28$ m
One observer estimates the angle of elevation to the basket of a hot air balloon to be $60^{\circ}$, while another observer $100$ m away estimates the angle of elevation to be $30^{\circ}$. Find : (a) The height of the basket from the ground. (b) The distance of the basket from the first observer's eye. (c) The horizontal distance of the second observer from the basket.
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Correct Figure Let $B$ is the basket of hot air balloon. $\tan 60^{\circ} = \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$ quad (i) $\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{h}{x + 100} \Rightarrow x = h\sqrt{3} - 100$ quad (ii) using (i) and (ii) (a) $h= (h\sqrt{3}-100)\sqrt{3} = 3h - 100\sqrt{3} \Rightarrow h = 50\sqrt{3}$ m (b) $\sin 60^{\circ} = \frac{\sqrt{3}}{2} = \frac{h}{y} \Rightarrow y = \frac{50\sqrt{3}}{\sqrt{3}/2} = 100$ m (c) $x = \frac{h}{\sqrt{3}} = 50$ m $\Rightarrow x+100=150$ m # ANOTHER SOLUTION AS PER BELOW FIGURE IS ALSO POSSIBLE Correct Figure Let $B$ is the basket of hot air balloon. $D$ and $C$ be the positions of the first and second observer's respectively. $\tan 60^{\circ} = \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$ quad (i) $\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{h}{100-x} \Rightarrow \sqrt{3}h = 100 - x$ quad (ii) (a) using (i) and (ii) $h = \sqrt{3} (100 - \sqrt{3}h) \Rightarrow h = 25\sqrt{3}$ m (b) $\sin 60^{\circ} = \frac{\sqrt{3}}{2} = \frac{h}{BD} \Rightarrow BD = 50$ m (c) $x = \frac{h}{\sqrt{3}} = 25$ m $AC = 100 - x = 75$ m
The angle of elevation of the top of a vertical tower from a point $P$ on the ground is $60^\circ$. From another point $Q$, $10$ m vertically above the first point $P$, its angle of elevation is $30^\circ$. Find : (a) The height of the tower. (b) The distance of the point $P$ from the foot of the tower. (c) The distance of the point $P$ from the top of the tower.
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Let $AD = h$ be the height of the tower and $AP = x$ be the distance of point $P$ from the foot of the tower. $Q$ is $10$ m vertically above $P$, so $PQ = 10$ m. Draw $QE \perp AD$. Then $AE = PQ = 10$ m and $ED = AD - AE = h - 10$. Also $QE = AP = x$. In right-angled $\triangle APD$: $\tan 60^\circ = \frac{AD}{AP} = \frac{h}{x}$ $\sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$ (1) In right-angled $\triangle QED$: $\tan 30^\circ = \frac{ED}{QE} = \frac{h-10}{x}$ $\frac{1}{\sqrt{3}} = \frac{h-10}{x} \Rightarrow x = \sqrt{3}(h-10)$ (2) Substitute $x$ from (1) into (2): $\frac{h}{\sqrt{3}} = \sqrt{3}(h-10)$ $h = 3(h-10)$ $h = 3h - 30$ $2h = 30 \Rightarrow h = 15$ m. (a) The height of the tower is $h = 15$ m. (b) From (1), $x = \frac{h}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3}$ m. The distance of the point $P$ from the foot of the tower is $5\sqrt{3}$ m. (c) The distance of the point $P$ from the top of the tower is $PD$. In right-angled $\triangle APD$: $\cos 60^\circ = \frac{AP}{PD} = \frac{x}{PD}$ $\frac{1}{2} = \frac{5\sqrt{3}}{PD}$ $PD = 2 \times 5\sqrt{3} = 10\sqrt{3}$ m. The distance of the point $P$ from the top of the tower is $10\sqrt{3}$ m.
Two pillars are standing on either side of a $80$ m wide road. Height of one pillar is $20$ m more than the height of the other pillar. From a point on the road between the pillars, the angle of elevation of the higher pillar is $60^{\circ}$, whereas that of the other pillar is $30^{\circ}$. Find the position of the point between the pillars and the height of each pillar. (Use $\sqrt{3} = 1.73$)
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Correct figure $1$ Mark Let $PQ$ and $RS$ be the pillars. $\tan 60^{\circ} = \sqrt{3} = \frac{h+20}{x} \implies h+20 = x\sqrt{3}$ (i) $\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{h}{80-x} \implies h\sqrt{3} = 80 - x$ (ii) Using (i) and (ii) $x = 28.65, h = 29.56$ $AP = 28.65$ m, $AR = 51.35$ m $PQ = h + 20 = 49.56$ m and $RS = 29.56$ m
A pole $6$m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is $60^\circ$ and the angle of depression of the point P from the top of the tower is $45^\circ$. Find the height of the tower and the distance of point P from the foot of the tower. (Use $\sqrt{3} = 1.73$)
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Sol. Correct figure Let BC be the pole and AB be the tower of height '$h$' m. $\tan 45^\circ = 1 = \frac{h}{x}$ $\Rightarrow h = x$ ----- (i) $\tan 60^\circ = \sqrt{3} = \frac{h+6}{x}$ $\Rightarrow h + 6 = x\sqrt{3}$ ----- (ii) Solving (i) & (ii) to get $h = 3 (\sqrt{3} + 1) = 8.19$ and $x = 8.19$ Therefore, the height of tower is $8.19$ m and the distance of point P from the foot of the tower is $8.19$ m
From the top of a building $60$ m high, the angles of depression of the top and bottom of the vertical lamp post are observed to be $30^\circ$ and $60^\circ$ respectively. (i) Find the horizontal distance between the building and the lamp post. (ii) Find the distance between the tops of the building and the lamp post.
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Correct figure Let AB be the building and CD be the lamp post of height '$h$' m. (i) $\tan 60^\circ = \frac{60}{x} = \sqrt{3} \Rightarrow x = 20\sqrt{3}$ m or AC = $20\sqrt{3}$ m (ii) $\cos 30^\circ = \frac{x}{BD} = \frac{\sqrt{3}}{2} \Rightarrow BD = \frac{2\times 20\sqrt{3}}{\sqrt{3}} = 40$ m
From the top of a $15$ m high building, the angle of elevation of the top of a tower is found to be $30^\circ$. From the bottom of the same building, the angle of elevation of the top of the tower is found to be $60^\circ$. Find the height of the tower and the distance between tower and the building.
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Correct figure Let CD be the building and AB be the tower. $\tan 30^\circ = \frac{h}{DE} = \frac{h}{x} = \frac{1}{\sqrt{3}}$ ----- (i) $\Rightarrow x = h\sqrt{3}$ $\tan 60^\circ = \frac{h+15}{x} = \sqrt{3}$ ----- (ii) $\Rightarrow h + 15 = x\sqrt{3}$ Solving (i) and (ii) to get $x = 7.5\sqrt{3}$ m or $\frac{15\sqrt{3}}{2}$ m and $h = \frac{15}{2} = 7.5$ m Hence height of the tower is $7.5 + 15 = 22.5$ m
From the top of a $45$ m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be $30^\circ$ and $60^\circ$. If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use $\sqrt{3} = 1.73$)
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1 mark for correct figure Let $AB$ be the light house and $C$ and $D$ be positions of ships. $\tan 60^\circ = \sqrt{3} = \frac{45}{y}$ $\Rightarrow y=15\sqrt{3}$ $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{45}{x}$ $\Rightarrow x = 45\sqrt{3}$ Distance between two ships $= x+y = 60\sqrt{3}$ $= 60 \times 1.73 = 103.8$ m
A person standing on the bank of a river observes that the angle of elevation of the top of a tower on the opposite bank is $60^\circ$. When he moves 30 m away from the bank, he finds the angle of elevation to be $30^\circ$. Find the height of the tower and width of the river. (Take $\sqrt{3} = 1.732$)
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1 mark for correct figure Let the height of tower BA be $h$ m and the width of river BC be $x$ m $\tan 60^\circ = \sqrt{3} = \frac{h}{x}$ $\Rightarrow h = \sqrt{3}x$ --- (i) $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{30+ x}$ $\Rightarrow h\sqrt{3} = 30 + x$ --- (ii) Solving (i) and (ii), we get $x = 15$ and $h = 15\sqrt{3}=15 \times 1.732 = 25.98$ m $\therefore$ Height of tower = 25.98 m and width of river = 15 m
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20 \text{ m}$ high building are $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.
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Correct Figure. In $\triangle BPA$ $\tan 45^\circ = 1 = \frac{20}{x}$ $\Rightarrow x = 20 \text{ m} \dots (i)$ Now, In $\triangle CPA$ $\tan 60^\circ = \sqrt{3} = \frac{h+20}{x}$ $\Rightarrow h + 20 = x\sqrt{3} \dots (ii)$ Solving (i) and (ii) $h = 20 (\sqrt{3} - 1) \text{ m}$ $\therefore$ Height of the tower is $20 (\sqrt{3} - 1) \text{ m}$.
Two pillars of equal lengths stand on either side of a road which is $100$ m wide, exactly opposite to each other. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are $60^\circ$ and $30^\circ$. Find the length of each pillar and distance of the point on the road from the pillars. (Use $\sqrt{3} = 1.732$)
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Correct figure Let $AB$ and $CD$ are two pillars of equal length $h$ m and let $P$ be the point on road $x$ m away from pillar $CD$. In $$\begin{aligned}& \triangle CDP \\ & \tan 60^\circ = \sqrt{3} = \frac{h}{x} \\ & \Rightarrow h = \sqrt{3} x ------(i) \\ & \text{In } \triangle ABP, \\ & \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{100-x} \\ & \Rightarrow h = \frac{100-x}{\sqrt{3}} -------(ii) \\ & \text{Solving eq.(i) and eq.(ii)} \\ & x = 25 \\ & \text{and } h = 25\sqrt{3} = 25 \times 1.732 = 43.3 \\ & \text{The length of each pillar is } 43.3 \text{ m and the distance of the point on the road from pillars is } 75 \text{ m and } 25 \text{ m respectively.}\end{aligned}$$
The angles of depression of the top and the bottom of a $50$ m high building from the top of a tower are $45^\circ$ and $60^\circ$, respectively. Find the height of the tower. (Use $\sqrt{3} = 1.73$)
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Correct figure Let AB be the tower of height H m and CD be the building. In $\triangle ABD$, $\tan 60^\circ = \sqrt{3} = \frac{H}{X}$ $\Rightarrow H = \sqrt{3} X$ --------(i) In $\triangle AEC$ $\tan 45^\circ = 1 = \frac{H-50}{X}$ $\Rightarrow X = H-50$ --------(ii) Solving equations (i) & (ii) $H = 25(3 + \sqrt{3}) = 25(3 + 1.73) = 118.25$ The height of tower is $118.25$ m
The angles of depression of the top and the bottom of a $8$ m tall building from the top of a multi-storeyed building are $30^\circ$ and $45^\circ$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings.
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Correct figure Let height of multi storeyed building AB be H m and CD is a tall building. Let the distance between the two buildings be X m. In $\Delta ABD$ $$\begin{aligned}& \tan 45^\circ = 1 = \frac{H}{X} \\ & \Rightarrow H = X \text{---------(i)} \\ & \text{In } \Delta AEC\end{aligned}$$ tan 30^ circ = frac{1}{ sqrt{3}} = frac{H-8}{X}$ \\ $X = sqrt{3} (H-8) ---------(ii) Solving equations (i) & (ii) $X = 4 (3 + \sqrt{3})$ and $H = 4 (3 + \sqrt{3})$ The height of the multi storeyed building is $4 (3 + \sqrt{3})$ m and the distance between the two buildings is $4 (3 + \sqrt{3})$ m.
From a point on a bridge across the river, the angles of depressions of the banks on opposite sides of the river are $30^{\circ}$ and $60^{\circ}$ respectively. If the bridge is at a height of $4 \text{ m}$ from the banks, find the width of the river.
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Correct Figure Let AB be the width of river In right $\triangle PAQ$, $\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{4}{x}$ $\Rightarrow 4\sqrt{3} = x$ In right $\triangle PBQ$, $\tan 60^{\circ} = \sqrt{3} = \frac{4}{y}$ $\Rightarrow y = \frac{4}{\sqrt{3}}$ Width of river = $x + y = 4\sqrt{3} + \frac{4}{\sqrt{3}} = \frac{16}{\sqrt{3}} \text{ m}$
From a window $15$ metres high above the ground in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are $30^{\circ}$ and $45^{\circ}$ respectively. Find the height of the opposite house. (Use $\sqrt{3} = 1.732$)
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For correct figure In right $\triangle BPC$, $\tan 45^{\circ} = 1 = \frac{15}{x}$ $\Rightarrow x = 15$ In right $\triangle BPD$, $\tan 30^{\circ} = \frac{h}{x}$ $h = \frac{15}{\sqrt{3}}$ or $5\sqrt{3}$ $\therefore$ Height of opposite house $(CD) = 5\sqrt{3} + 15$ $= 5 (1.732) + 15 = 23.66$ m
Two poles of equal height are standing opposite to each other, on either side of a road, which is $100$ m wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^{\circ}$ and $30^{\circ}$ respectively. Find the height of the poles.
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Let 'h' be the height of pole. $\frac{h}{x} = \tan 60^{\circ} = \sqrt{3}$ $\Rightarrow h = \sqrt{3}x$ --- (1) Also, $\frac{h}{100-x} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ $\Rightarrow h = \frac{100 - x}{\sqrt{3}}$ --- (2) From (1) and (2), we get $x = 25$ Hence, $h = 25\sqrt{3}$ m
Two ships are sailing in the sea on either side of a lighthouse. The angles of depression to the two ships as observed from the top of the lighthouse are $60^\circ$ and $45^\circ$, respectively. If the distance between the ships is $100 \left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$ m, then find the height of the lighthouse.
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Correct figure Here, AB represents the height of the lighthouse. In right $\triangle ABP$ $\frac{AB}{PB} = \tan 60^\circ = \sqrt{3}$ $\Rightarrow PB = \frac{AB}{\sqrt{3}}$ ----- (1) In right $\triangle ABQ$ $\frac{AB}{BQ} = \tan 45^\circ = 1$ $\Rightarrow BQ = AB$ ----- (2) Adding (1) and (2), we have $PB + BQ = \frac{AB}{\sqrt{3}} + AB$ $\Rightarrow PQ = AB \left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$ $100 \left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) = AB \left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$ $\Rightarrow AB = 100$ m
The angles of depression of the top and the bottom of an $8$ m tall building from the top of another multistoried building are $30^\circ$ and $45^\circ$, respectively. Find the height of the multistoried building and the distance between the two buildings.
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Correct figure Here, CD represents the multistoried building. In right $\triangle DEA$ $\frac{DE}{AE} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ $\Rightarrow AE = \sqrt{3} DE$ ----- (1) In right $\triangle DCB$ $\frac{DC}{BC} = \tan 45^\circ = 1$ $\Rightarrow BC = DC$ ----- (2) From figure, AE = BC $\therefore \sqrt{3} DE = DC$ $\Rightarrow \sqrt{3} (DC - 8) = DC$ $\Rightarrow DC = \frac{8\sqrt{3}}{\sqrt{3}-1}$ $= \frac{8\sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = (12 + 4\sqrt{3})$ m From (2), BC = $(12 + 4\sqrt{3})$ m
A girl $1.5$ m tall is standing at some distance from a $30$ m high tower. The angle of elevation from her eye to the top of the tower increases from $30^\circ$ to $60^\circ$ as she walks towards the tower. Find the distance she walked towards the tower.
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Correct figure ($1$ mark) $AF = 30 - 1.5 = 28.5$ m ($1/2$ mark) In right $\triangle AFE$ $\frac{28.5}{EF} = \tan 60^\circ = \sqrt{3}$ $\Rightarrow EF = \frac{28.5}{\sqrt{3}}$ m ($1$ mark) In right $\triangle ADF$ $\frac{28.5}{DF} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ $\Rightarrow DF = 28.5 \sqrt{3}$ m ($1$ mark) Distance travelled by the girl towards the tower, $DE = DF - EF$ $= 28.5 \sqrt{3} - \frac{28.5}{\sqrt{3}}$ ($1$ mark) $= \frac{57}{\sqrt{3}} = 19 \sqrt{3}$ m or $32.91$ m approx. ($1/2$ mark) Distance travelled by girl towards the tower is $19 \sqrt{3}$ m or $32.91$ m.
A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be $30^{\circ}$ and $45^{\circ}$ as it moves from point P to point Q. The height of the lighthouse is $50$ metres. Based on the information given above, answer the following questions: (i) Find the distance of the ship from the base of the lighthouse when it is at point Q, where the angle of depression is $45^{\circ}$. (ii) Find the measures of $\angle PBA$ and $\angle QBA$. (iii) (a) Find the distance travelled by the ship between points P and Q. OR (b) If the ship continues moving towards the shore and takes $10$ minutes to travel from Q to A, calculate the speed of the ship in $\operatorname{km/h}$, from Q to A.
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Sol. (i) $\angle AQB = \angle QBX = 45^{\circ}$ and $\angle APB = \angle PBX = 30^{\circ}$ In $\triangle AQB$, $\tan 45^{\circ} = \frac{50}{AQ}$ $AQ = 50 \operatorname{m}$ (ii) $\angle PBA = 60^{\circ}$ $\angle QBA = 45^{\circ}$ (iii)(a) In $\triangle APB$, $\tan 30^{\circ} = \frac{50}{AP}$ $AP = 50\sqrt{3} \operatorname{m}$ Distance travelled by the ship $= PQ = 50\sqrt{3} - 50 = 50(\sqrt{3} - 1) \operatorname{m}$ or $36.5 \operatorname{m}$ OR (iii)(b) Speed of the ship $= \frac{50 \operatorname{metres}}{10 \operatorname{minutes}}$ $= 0.3 \operatorname{km/h}$
A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there was an accident. The ambulance was travelling at the speed of $60$ km/h. The drone stopped at a point Q, $100$ m vertically above the point P. The angle of depression of the ambulance was found to be $30^\circ$ at a particular instant. Based on above information, answer the following questions : (i) Represent the above situation with the help of a diagram. (ii) Find the distance between the ambulance and the site of accident (P) at the particular instant. (Use $\sqrt{3}= 1.73$) (iii) (a) Find the time (in seconds) in which the angle of depression changes from $30^\circ$ to $45^\circ$. OR (iii) (b) How long (in seconds) will the ambulance take to reach point P from a point T on the highway such that angle of depression of the ambulance at T is $60^\circ$ from the drone ?
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(i) For correct figure (ii) In $\triangle PQR$, $\frac{100}{d} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ $\Rightarrow d = 100\sqrt{3} = 173$ m (iii) (a) For correct figure In $\triangle PQM$, $\frac{100}{173-x} = \tan 45^\circ = 1$ $\Rightarrow x = 73$ m Time taken = $\frac{73\times 18}{60\times5} = \frac{219}{50}$ or $4.4$ seconds (approx.) OR (iii) (b) For correct figure In $\triangle PQT$, $\frac{100}{y} = \tan 60^\circ = \sqrt{3}$ $\Rightarrow y = \frac{100}{\sqrt{3}} = \frac{100\sqrt{3}}{3}$ or $173/3$ m Time taken = $\frac{100\sqrt{3} \times 18}{3 \times 60 \times5} = 2\sqrt{3}$ or $3.5$ seconds (approx.)
The angle of elevation of an aeroplane from a point on the ground is $45^\circ$. After a flight of $15$ seconds, the elevation changes to $30^\circ$. If the aeroplane is flying at a constant height of $3000$ meters, find the speed of the aeroplane in km/h. [Take $\sqrt{3} = 1.732$]
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Let the height of the aeroplane be $h = 3000$ m. Let the initial position of the aeroplane be A and final position be B. Let the point on the ground be P. In $\triangle APM$, $\tan 45^\circ = \frac{AM}{PM} = \frac{3000}{x}$ $1 = \frac{3000}{x} \Rightarrow x = 3000$ m. In $\triangle BPN$, $\tan 30^\circ = \frac{BN}{PN} = \frac{3000}{x+y}$ $\frac{1}{\sqrt{3}} = \frac{3000}{3000+y} \Rightarrow 3000+y = 3000\sqrt{3}$ $y = 3000\sqrt{3} - 3000 = 3000(\sqrt{3}-1)$ $y = 3000(1.732-1) = 3000(0.732) = 2196$ m. Distance covered by aeroplane in $15$ seconds is $y = 2196$ m. Speed of aeroplane $= \frac{\text{Distance}}{\text{Time}} = \frac{2196 \text{ m}}{15 \text{ s}}$ Speed $= \frac{2196}{15}$ m/s $= 146.4$ m/s. To convert to km/h: $146.4 \times \frac{3600}{1000} = 146.4 \times 3.6 = 527.04$ km/h.
The angle of elevation of an aircraft from a point A on the ground is $60^\circ$. After a flight of $30$ seconds, the angle of elevation changes to $30^\circ$. The aircraft is flying at a constant height of $3500\sqrt{3}$ m at a uniform speed. Find the speed of the aircraft.
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1 mark for correct figure Let P and Q be the positions of aircraft at two different times. $\tan 60^\circ = \sqrt{3} = \frac{3500\sqrt{3}}{x}$ (1) $\Rightarrow x = 3500$ m ... (i) ($\frac{1}{2}$) $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{3500\sqrt{3}}{x+y} \Rightarrow x+y = 10500$ (1) Using (i), we get $y = 7000$ ($\frac{1}{2}$) $\therefore$ Speed of aircraft $= \frac{7000}{30}$ or $233. 3$ m/s approx. (1)
The angle of elevation of a jet plane from a point A on the ground is $60^\circ$. After a flight of $30$ seconds, the angle of elevation changes to $30^\circ$. If the jet plane is flying at a constant height of $3600 \sqrt{3} \text{ m}$, find the speed of the jet plane.
A man on a cliff observes a boat at an angle of depression of $30^{\circ}$ which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be $60^{\circ}$. Find the time taken by the boat from here to reach the shore.
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Sol. Correct fig. In $\triangle$QAB $\tan60^{\circ} = \sqrt{3} = \frac{h}{x}$ $h=\sqrt{3}x$.........(i) In $\triangle$PAB $\tan30^{\circ} = \frac{h}{y+x} = \frac{1}{\sqrt{3}}$ $y + x=\sqrt{3}h$.........(ii) solving (i) and (ii) $y=2x$ Time taken by the boat from Q to A = $\frac{1}{2} \times 6 =3$ min.
The angle of elevation of an airborne helicopter from a point $A$ on the ground is $45^\circ$. After a flight of $15$ seconds, the angle of elevation of the helicopter changes to $30^\circ$. If the helicopter is flying at a constant height of $2000$ m, find the speed of the helicopter. (Take $\sqrt{3} = 1.732$)
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Correct figure ($1$ mark) In right $\triangle APB$ $\frac{2000}{AP} = \tan 45^\circ = 1$ $\Rightarrow AP = 2000$ m ($1/2$ mark) In right $\triangle ACD$ $\frac{2000}{AC} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ $\Rightarrow AC = 2000 \sqrt{3}$ m ($1/2$ mark) $BD = PC = AC - AP = 2000 \sqrt{3} - 2000$ ($1$ mark) $= 2000 (1.732 - 1)$ $= 1464$ m ($1$ mark) Time taken from $B$ to $D = 15$ sec Speed $= \frac{1464}{15} = 97.6$ m/s ($1$ mark)
If a vertical pole of length $7.5 \text{ m}$ casts a shadow $5 \text{ m}$ long on the ground and at the same time, a tower casts a shadow $24 \text{ m}$ long, then the height of the tower is :
A golf ball is spherical with about $300 - 500$ dimples that help increase its velocity while in play. Golf balls are traditionally white but available in colours also. In the given figure, a golf ball has diameter $4.2$ cm and the surface has $315$ dimples (hemi-spherical) of radius $2$ mm. Based on the above, answer the following questions : (i) Find the surface area of one such dimple. (ii) Find the volume of the material dug out to make one dimple. (iii) (a) Find the total surface area exposed to the surroundings. OR (iii) (b) Find the volume of the golf ball.
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(i) $SA = 2\pi r^2 = 2 \times \frac{22}{7} \times 4 = \frac{176}{7} \text{ mm}^2$ or $25.1 \text{ mm}^2$ (ii) Volume of material dug out to make one dimple $= \frac{2}{3} \times \frac{22}{7} \times 8 = \frac{352}{21} \text{ mm}^3$ or $16.76 \text{ mm}^3$ (iii)(a) radius of ball $= 21$ mm Total surface area exposed to surroundings $= 4\pi(21)^2 - 315 \times \pi(2)^2 + 315 \times 2\pi(2)^2$ $= 4 \times \frac{22}{7} \times 21 \times 21 + \frac{22}{7} \times 315 \times 4$ $= 9504 \text{ mm}^2$ OR (iii) (b) Volume of the golf ball $= \frac{4}{3}\pi(21)^3 - 315 \times \frac{2}{3}\pi(2)^3$ $= 33528 \text{ mm}^3$
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