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Case Study - 2
An injured bird was found on the roof of a building. The building is $15$ m high. A fireman was called to rescue the bird. The fireman used an adjustable ladder to reach the roof. He placed the ladder in such a way that the ladder makes an angle of $60^{\circ}$ with the ground in order to reach the roof.
Based on the above information, answer the following questions :
(i) Find the length of the ladder used by the fireman to reach the roof.
(ii) Find the distance of the point on the ground at which the ladder was fixed from the bottom of the building.
(iii) In order to avoid skidding, the fireman placed the ladder in such a way that the bottom of the ladder touches the base of the wall which is opposite to the building, making an angle of $30^{\circ}$ with the ground.
(a) Draw a neat diagram to represent the above situation and hence find the width of the road between the building and the wall.
OR
(b) Find the length of the ladder used by the fireman in this case.
An injured bird was found on the roof of a building. The building is $15$ m high. A fireman was called to rescue the bird. The fireman used an adjustable ladder to reach the roof. He placed the ladder in such a way that the ladder makes an angle of $60^{\circ}$ with the ground in order to reach the roof.
Based on the above information, answer the following questions :
(i) Find the length of the ladder used by the fireman to reach the roof.
(ii) Find the distance of the point on the ground at which the ladder was fixed from the bottom of the building.
(iii) In order to avoid skidding, the fireman placed the ladder in such a way that the bottom of the ladder touches the base of the wall which is opposite to the building, making an angle of $30^{\circ}$ with the ground.
(a) Draw a neat diagram to represent the above situation and hence find the width of the road between the building and the wall.
OR
(b) Find the length of the ladder used by the fireman in this case.
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Solution: (i) Let the length of the ladder be '$a$'
$\frac{15}{a} = \sin 60^{\circ} \Rightarrow a = \frac{30}{\sqrt{3}}$ or $10\sqrt{3}$
Thus the length of the ladder is $\frac{30}{\sqrt{3}}$ m or $10\sqrt{3}$ m
(ii) Let the distance of the point on the ground be '$x$'
$\frac{15}{x} = \tan 60^{\circ} \Rightarrow x = \frac{15}{\sqrt{3}}$ or $5\sqrt{3}$
Thus, the distance of the point on the ground is $\frac{15}{\sqrt{3}}$ m or $5\sqrt{3}$ m
(iii) (a) Let the width of the road be $y$
$\frac{15}{y} = \tan 30^{\circ} \Rightarrow y = 15\sqrt{3}$
Thus, the width of the road is $15\sqrt{3}$ m.
OR
(b) Let the length of the ladder be $l$.
$\frac{15}{l} = \sin 30^{\circ} \Rightarrow l = 30$
Thus, the length of the ladder is $30$ m.
$\frac{15}{a} = \sin 60^{\circ} \Rightarrow a = \frac{30}{\sqrt{3}}$ or $10\sqrt{3}$
Thus the length of the ladder is $\frac{30}{\sqrt{3}}$ m or $10\sqrt{3}$ m
(ii) Let the distance of the point on the ground be '$x$'
$\frac{15}{x} = \tan 60^{\circ} \Rightarrow x = \frac{15}{\sqrt{3}}$ or $5\sqrt{3}$
Thus, the distance of the point on the ground is $\frac{15}{\sqrt{3}}$ m or $5\sqrt{3}$ m
(iii) (a) Let the width of the road be $y$
$\frac{15}{y} = \tan 30^{\circ} \Rightarrow y = 15\sqrt{3}$
Thus, the width of the road is $15\sqrt{3}$ m.
OR
(b) Let the length of the ladder be $l$.
$\frac{15}{l} = \sin 30^{\circ} \Rightarrow l = 30$
Thus, the length of the ladder is $30$ m.