Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 49

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Question 1
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If a flag-staff of $$6m$$ height placed on the top of a tower throws a shadow of $$2\sqrt { 3 } m$$ along the ground, then what is the angle that the sun makes with the ground?

A
$${ 60 }^{ o }$$

B
$${ 45 }^{ o }\quad $$

C
$${ 30 }^{ o }$$

D
$${ 15 }^{ o }$$

Solution

Let the Angle be $$x$$
and Length of shadow$$=2\sqrt3 m$$
Length of Tower $$=6m$$

So, $$\tan x=\dfrac{6}{2\sqrt3}\Rightarrow \tan x=\sqrt3\Rightarrow \angle x=60^o$$

Hence, the answer is $$60^{o}$$.
Question 2
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The angle of elevation of the top of a tower from the top and bottom of a building of height $$a$$ are $$30^{\circ}$$ and $$45^{\circ}$$ respectively. If the tower and the building are at same level, the height of the tower is

A
$$\dfrac{\sqrt{3}a(1+\sqrt{3})}{2}$$

B
$$a\left(\sqrt{3}+1\right)$$

C
$$\sqrt{3}a$$

D
$$a\left(\sqrt{3}-1\right)$$

Solution
Question 3
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The angle of elevation of the top of pole from the point $$D$$ on the ground be $$\alpha$$, whereas the angle of elevation of the top of pole from the point which is at a height $$'b'$$ above point $$D$$ is $$\beta$$, then the length of pole is:

A
$$\dfrac { b\cot \beta \tan \alpha }{ \cot \beta \tan \alpha -1 }$$

B
$$b \cot \alpha. \tan\beta$$

C
$$b \ \tan \alpha . \tan \beta$$

D
$$b \ \cot \alpha . \cot \beta$$

Solution

Let $$AB$$ be the Pole. C be the point at a height 'b' above point D.
Also, $$DC=b=AE, AD = CE$$, {from geometry}
Let say $$BE=x$$
In $$\triangle BCE,$$

$$ \dfrac{CE}{x}=\cot \beta $$

$$ CE=x \cot \beta = AD $$ .........(1)

In $$\triangle ADB,$$

$$\tan \alpha =\dfrac { AB }{ AD } =\dfrac { AE+BE }{ AD } =\dfrac { x+b }{ x\cot \beta } $$ {From (1)}

$$(\tan \alpha) x\cot \beta =x+b$$

$$x=\dfrac { b }{ \cot \beta \tan \alpha -1 } $$

$$AB=x+b=\dfrac { b\cot \beta \tan \alpha }{ \cot \beta \tan \alpha -1 } $$
Question 4
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The angle of elevation of the top of an unfinished tower at a point $$120m$$ from its base is $$45^o$$. How much higher must the tower be raised so that its angle of elevation becomes $$60^o$$ at the same point?

A
$$90m$$

B
$$92m$$

C
$$97m$$

D
$$87.84m$$

Solution

Let PA be the unfinished tower.
Let B be the point of observation i.e. 120 m away from the base of the tower.
Now, AB = 120m
Let, $$\angle{ABP} = 45°$$
Let h m be the height by which the unfinished tower be raised such that its angle of elevation of the top from the same point becomes 60°.
Let CA = h $$\& \; \angle{ABC} = 60°$$
In triangle ABP,
$$\tan{45°} = \cfrac{PA}{AB}$$
$$\Rightarrow \; 1 = \cfrac{PA}{120}$$
$$\Rightarrow \; PA = 120M$$
Now, in triangle ABC,
$$\tan{60°} = \cfrac{CA}{AB}$$
$$\sqrt{3} = \cfrac{120 + h}{120}$$
$$h + 120 = 120\sqrt{3}$$
$$h = 120 (\sqrt{3} - 1)$$
$$h = 120 (1.732 - 1) \; \rightarrow (as \; \sqrt{3} = 1.732)$$
$$h = 120 \times 0.732$$
$$h = 87.84m$$
Question 5
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$$ABCD$$ is a square plot. The angle of elevation of the top of a pole standing at $$D$$ from $$A$$ or $$C$$ is $$30^\circ$$ and that from $$B$$ is $$\theta$$, then $$\tan \theta$$ is equal to

A
$$\sqrt{6}$$

B
$$\dfrac{1}{\sqrt{6}}$$

C
$$\dfrac{1}{\sqrt{3}}$$

D
$$\dfrac{2}{\sqrt{6}}$$

Solution

Let the height of the pole be $$h$$ m

Then $$h = a \tan 30^\circ$$

$$= a \sqrt{3}$$ $$\bigg[$$from $$\triangle PAD$$ or $$\triangle PCD \bigg]$$

$$h = BD \times \tan \theta$$

$$= \sqrt{(a^2 + a^2)}\,\tan \theta$$

$$=a\sqrt{2} \tan \theta$$

From $$\triangle PBD,$$ diagonal $$BD = a \sqrt{2}$$

or $$\dfrac{a}{\sqrt{3}} = a\sqrt{2} \,\tan\theta$$

$$\tan \theta =\dfrac{1}{\sqrt{6}}$$
Question 6
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From the top of a cliff of height $$a$$, the angle of depression of the foot of a certain tower is found to be double the angle of elevation of the top of the tower of height $$h$$. If $$\theta$$ be the angle of elevation then its value is:

A
$$\cos^{-1} \sqrt{\dfrac{2h}{a}}$$

B
$$\sin^{-1} \sqrt{\dfrac{2h}{a}}$$

C
$$\sin^{-1} \sqrt{\dfrac{a}{2 - h}}$$

D
$$\tan^{-1} \sqrt{3 - \dfrac{2h}{a}}$$

Solution

$$\textbf{ Step 1 : Apply trigonometric ratio in right angled trinagle }$$

$$\text{Let AB be the tower and PQ be the cliff}$$

$$\text{In}$$$$\triangle QBM$$

$$\Rightarrow h - a = x \tan \theta..... (1)$$

$$\text{In}$$ $$\triangle QAP$$

$$\Rightarrow a = x \tan 2\theta........(2)$$

$$\text{On dividing equ (1) by (2), we get}$$

$$\Rightarrow \dfrac{h-a}{a} = \dfrac{\tan \theta}{\tan 2\theta}$$

$$\textbf{ Step 2 : Apply the trigononometric identity }$$

$$\boldsymbol{\tan 2\theta =\dfrac{2\tan \theta}{1-\tan ^2 \theta}}$$

$$\Rightarrow \dfrac{h}{a} -1 = \dfrac{1-\tan ^2 \theta}{2}$$

$$\Rightarrow 1-\tan ^2 \theta = \dfrac{2h}{a} -2$$

$$\Rightarrow \tan ^2 \theta =3-\dfrac{2h}{a}$$

$$\Rightarrow \tan \theta = \sqrt{3-\dfrac{2h}{a}}$$

$$\Rightarrow \theta = \tan ^{-1} \left ( \sqrt{3-\dfrac{2h}{a}} \right )$$
Question 7
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An observer on the top of a tree finds the angle of depression of a car moving towards the tree to be $${30^ \circ }$$. After $$3\text{ min.}$$ this angle becomes $${60^ \circ }$$. After how much more time will the car reach the tree

A
$$4\text{ min.}$$

B
$$1.5\text{ min.}$$

C
$$4.5\text{ min.}$$

D
$$2\text{ min.}$$

Solution

Let the speed of the car be $$x \text{ units/min}$$ and the height of the tree equal to $$h$$.
So, $$CD$$ will be equal to $$3x$$.

From the above figure, we can conclude that, $$\angle ACB=60^\circ$$ and $$\angle ADB=30^\circ.$$

In $$\triangle ABC,$$
$$\tan { 60^\circ } =\cfrac { h }{ BC } \\ \quad\;\;\;\;\;\;h=\sqrt { 3 } BC$$

In $$\triangle ABD,$$
$$\tan { 30^\circ } =\cfrac { h }{ BD } \\ \quad\;\sqrt3 h=BD$$

Also,
$$\begin{aligned}{}BD &= \sqrt 3 h\\BC + CD& = \sqrt 3 h\\\frac{h}{{\sqrt 3 }} + 3x &= \sqrt 3 h\\h + 3\sqrt 3 x &= 3h\\2h &= 3\sqrt 3 x\\h &= \frac{{3\sqrt 3 x}}{2}\end{aligned}$$

So, the time taken by car to reach the tower will be,
$$\begin{aligned}{} \text{Time}&=\frac{\text{Distance}}{\text{Speed}}\\&= \frac{{BC}}{x}\\& = \frac{{\frac{h}{{\sqrt 3 }}}}{x}\\ &= \frac{h}{{\sqrt 3 x}}\\ &= \frac{{3\sqrt 3 x}}{2} \times \frac{1}{{\sqrt 3 x}}\\& = 1.5\end{aligned}$$

So, the time taken by car to reach the base of the tree is equal to $$1.5 \text{ min.}$$
Question 8
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A uniform rod is rested on a wall and its lower end is tied to the wall with the help of a horizontal string of negligible mass as shown. Length of rod is $$L$$ and height of the wall is $$h$$. What will be tension in the string? (all surface are frictionless)

A
$$\dfrac { mg\ell\sin { \theta } }{ 2h\cos { \theta } }$$

B
$$\dfrac { mg\ell\cos ^{ 2 }{ \theta } }{ 2h\sin { \theta } }$$

C
$$\dfrac { mg\ell\sin { \theta } \cos ^{ 2 }{ \theta } }{ 2h }$$

D
$$\dfrac { mg\ell }{ 2h\sin { \theta } \cos ^{ 2 }{ \theta } }$$

Solution
Question 9
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From the top of a $$25\ \text{m}$$ high pillar at the top of tower, angle of elevation is same as the angle of depression of foot of tower, then height of tower is:

A
$$25\ \text{m}$$

B
$$100\ \text{m}$$

C
$$75\ \text{m}$$

D
$$50\ \text{m}$$

Solution

$$OB$$ is the length of the pillar and $$AD$$ is the length of the tower.
Let $$\theta$$ be angle of elevation to top of tower and angle of depression to foot of tower.
Height of pillar$$=h=25\ \text{m}$$
Height of tower$$=h+d$$
$$\tan \theta=\cfrac{h}{BC}=\cfrac{d}{BC}$$
$$\therefore h=d$$
$$\therefore$$ Height of tower $$=h+d=2h=50\ \text{m}$$
Question 10
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A ship $$A$$ is moving westward with a speed of $$10$$ km/s. and ship $$B$$ $$100$$ km south of $$A$$ is moving northward with a speed of $$10$$ km/hr. The time after which distance between then becomes shortest is:

A
$$5$$ hr

B
$$5\sqrt{2}$$ hr

C
$$10\sqrt{2}$$ hr

D
$$0$$ hr

Solution

According to the problem
In diagram, the shortest distance b/w the ship A and B is QM.
$$\sin {45^0} = \frac{{MQ}}{{OQ}}$$
Implies that
Hence Also,
$$VAB = \sqrt {\left( {{V_A}^2 - {V_B}^2} \right)} = \sqrt {{{10}^2} + {{10}^2}} 10\sqrt 2 km/hr$$
For $$t$$,
$$t = \dfrac{{MQ}}{{VAB}} = \dfrac{{50\sqrt 2 }}{{10\sqrt 2 }} = 5hr$$