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Jigyasa is fly fishing in a stream sitting at the height of $3.1$ m above the surface of water. The fishing rod is $2\sqrt{2}$ m long and the fly at the end of the string resting on the surface of water is $5$ m away from the base of the platform on which Jigyasa is sitting.
The fishing rod makes an angle of $45^\circ$ with the horizontal and the string which is fully taut makes an angle of $60^\circ$ with the horizontal as shown in the given figure.
Based on the above information, answer the following questions :
(Use $\sqrt{3} = 1.7$)
(i) What is the height of the top of the fishing rod above the surface of water?
(ii) How far is the fly from the point on the surface of water directly below the top of the fishing rod ?
(iii) (a) Find the length of the string when it is fully taut.
OR
(iii) (b) A fish pulled the string such that it makes $30^\circ$ angle with the horizontal now. What is the length of the string in this case?
The fishing rod makes an angle of $45^\circ$ with the horizontal and the string which is fully taut makes an angle of $60^\circ$ with the horizontal as shown in the given figure.
Based on the above information, answer the following questions :
(Use $\sqrt{3} = 1.7$)
(i) What is the height of the top of the fishing rod above the surface of water?
(ii) How far is the fly from the point on the surface of water directly below the top of the fishing rod ?
(iii) (a) Find the length of the string when it is fully taut.
OR
(iii) (b) A fish pulled the string such that it makes $30^\circ$ angle with the horizontal now. What is the length of the string in this case?
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Solution
(i) In $\Delta AFD$,
$\sin 45^\circ = \frac{h - 3.1}{2\sqrt{2}}$ gives $h = 5.1$ m (1 Mark)
(ii) In $\Delta ACB$,
$\tan 60^\circ = \frac{h}{BC}$ gives $BC = \frac{5.1}{1.7} = 3$ m (1 Mark)
(iii) (a) Let AB be the length of the string
In $\Delta ACB$,
$\sin 60^\circ = \frac{h}{AB}$ (1 Mark)
$\frac{\sqrt{3}}{2} = \frac{5.1}{AB}$
$\Rightarrow AB = 6$ m (1 Mark)
OR
(iii) (b)
$\sin 30^\circ = \frac{5.1}{l}$ (1 Mark)
$\frac{1}{2} = \frac{5.1}{l}$
$l = 10.2$ m (1 Mark)
(i) In $\Delta AFD$,
$\sin 45^\circ = \frac{h - 3.1}{2\sqrt{2}}$ gives $h = 5.1$ m (1 Mark)
(ii) In $\Delta ACB$,
$\tan 60^\circ = \frac{h}{BC}$ gives $BC = \frac{5.1}{1.7} = 3$ m (1 Mark)
(iii) (a) Let AB be the length of the string
In $\Delta ACB$,
$\sin 60^\circ = \frac{h}{AB}$ (1 Mark)
$\frac{\sqrt{3}}{2} = \frac{5.1}{AB}$
$\Rightarrow AB = 6$ m (1 Mark)
OR
(iii) (b)
$\sin 30^\circ = \frac{5.1}{l}$ (1 Mark)
$\frac{1}{2} = \frac{5.1}{l}$
$l = 10.2$ m (1 Mark)