Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 55

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Question 1
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If ratio of length of a vertical rod and length of its shadow is $$1 : \sqrt{3}$$, then angle of elevation of sun is :

A
$$30^{0}$$

B
$$45^{0}$$

C
$$60^{0}$$

D
$$90^{0}$$

Solution

Let length of vertical rod is $$BC$$ and length of shadow is $$AB$$.
Let angle of elevation of sun is $$\theta$$ then,
From right $$\Delta ABC$$,
$$\tan \theta =\dfrac{BC}{AB}$$
$$=\dfrac{1}{\sqrt{3}} = \tan 30^{0}$$
or $$\theta = 30^{0}$$
Hence, angle of elevation of sun $$= 30^{0}$$
So, correct choice is (A).
Question 2
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A kite is flying at a height of $$30 m$$ from the ground. The length of string from the kite to the ground is $$60 m$$. Assuming that there is no slack in the string, the angle of elevation of the kite at the
ground is :

A
$$45^{0}$$

B
$$30^{0}$$

C
$$60^{0}$$

D
$$90^{0}$$

Solution

Let $$A$$ is a kite which is $$30$$ meter high from the ground which is flying with string of $$60 m$$ long.
Let $$\angle ACB = \theta$$
From right angled
$$\sin \theta =\dfrac{AB}{AC}$$
$$\sin \theta =\dfrac{30}{60}$$
$$\sin \theta =\dfrac{1}{2}$$
$$sin \theta = \sin 30^{0}$$
$$\theta = 30^{0}$$
So, correct choice is (B)
Question 3
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If the height of a pole is $$6 m$$ and the length of its shadow is $$2\sqrt{3} m$$, then the angle of elevation of sun is equal to

A
$$60^{0}$$

B
$$45^{0}$$

C
$$30^{0}$$

D
$$90^{0}$$

Solution

Let $$AB$$ is a pole of height $$6 m$$
Length of a shadow of pole $$BC = 2\sqrt{3} m$$
Let angle of elevation of sun $$\angle ACB = \theta$$
From right angled $$\Delta ABC$$,
$$\tan \theta =\frac{AB}{BC}$$
$$\tan \theta =\frac{6}{2\sqrt{3}}$$
$$\tan \theta = \sqrt{3}$$
$$\tan \theta = \tan 60^{0}$$
$$\theta = 60^{0}$$
So, correct choice is (A).
Question 4
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A man on the top of the light house observes a boat comming towards it. lf it takes $$10$$ minutes forthe angle of depresion to change from $$30^{0}$$ to $$45^{0}$$, the time in minutes taken by the boat to reach the shore is

A
$$13.66$$

B
$$23.66$$

C
$$33.66$$

D
$$44.66$$

Solution

Let the height of light house be $$h$$
Let the point at which angle of depression is $$30^{0}$$ is $$A$$ and the point at which angle of depression is $$45^{0}$$ is $$B$$
Let the foot of light house be $$o$$
Let $$OB=x$$ and $$AB=y$$

We have $$tan(30)=\dfrac{1}{\sqrt3}=\dfrac{h}{x+y}$$ and $$tan(45)=1=\dfrac{h}{x}$$

So we get $$y=(\sqrt3-1)x$$

Let the speed of boat be $$v$$
Given $$y=v \times 10$$ , Let the time taken by boat to travel distance $$OA$$ be $$t$$

Therefore,

$$OA=x+y=vt$$
$$\Rightarrow v\times 10(1+\dfrac{1}{\sqrt3-1})=vt$$

$$\Rightarrow t=5(3+\sqrt3)=23.66$$
Question 5
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A flag staff stand in the centre of a rectangular field whose diagonal is $$1200$$ $$\text{m}$$, and subtends angles $$15^{\circ}$$ and $$45^{\circ}$$ at the mid points of the sides of the field. The height of the flag staff is

A
$$200\text{ m}$$

B
$${300}(\sqrt{2+\sqrt{3}})\text{ m}$$

C
$$300(\sqrt{2-\sqrt{3}})\text{ m}$$

D
$$400\text{ m}$$

Solution

Let ABCD be a rectangular field with center at O. Let E and F are the mid-points of sides AD and DC respectively.
Let OP be the flagstaff of height h .
In right $$\triangle OPE$$
$$\tan 15^{\circ}=\displaystyle \frac{h}{OE}$$
$$\Rightarrow OE=h\cot 15^{\circ}$$
$$\Rightarrow OE=h(2+\sqrt{3})$$ .....(1) $$[\because \cot 15^{\circ}=2+\sqrt 3]$$

In right $$\triangle OPF$$
$$\tan 45^{\circ}=\displaystyle \frac{h}{OF}$$
$$\Rightarrow OF=h\cot 45^{\circ}$$
$$\Rightarrow OF=h$$ .....(2)

In right $$\triangle EOF$$
$$EF^{2}=OE^{2}+OF^{2}$$
$$\Rightarrow EF^{2}=h^{2}(1+(2+\sqrt{3})^2) $$ (by (1) and (2))
$$=h^{2}(8+4\sqrt{3})$$
$$\Rightarrow EF=2h\sqrt{2+\sqrt{3}}$$ ....(3)

Now, $$EF= \displaystyle \frac{1}{2}AC$$
$$\Rightarrow EF=600$$
So, by (3), we get
$$h=\displaystyle \frac{300}{\sqrt{2+\sqrt{3}}}$$
$$\Rightarrow h=(300 \sqrt{2-\sqrt{3}}) m$$
Question 6
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The angle of elevation of the top of a vertical pole when observed from each vertex of a regular hexagon is $$\displaystyle \frac{\pi}{3}$$ if the area of the circle circum scribing the hexagon be $$A\ m$$, then the height of the tower is

A
$$\displaystyle \frac{2A}{\sqrt{3\pi}}\ m$$

B
$$\displaystyle \frac{A}{\sqrt{3\pi}} \ m$$

C
$$2\sqrt{\dfrac{A}{3\pi}}\ m$$

D
$$\sqrt{\dfrac{3A}{\pi}}\ m$$

Solution

Given that:
Angle of elevation from vertex of a regular hexagon$$=\cfrac{\pi}{3}$$
Area of the circle circumscribing the hexagon $$=Am$$
To find:
Height of the tower $$=?$$
Solution:
Let $$r$$ be the radius of circle and $$h$$ be the height of tower.
Area of circle $$=\pi r^2$$
$$\Rightarrow A=\pi r^2$$
$$\Rightarrow r=\sqrt{\cfrac{A}{\pi}}$$
Now, $$\tan \left(\cfrac{\pi}{3}\right)=\cfrac hr$$
or, $$h=r\tan\left(\cfrac{\pi}{3}\right)$$
or, $$h=\sqrt{\cfrac{A}{\pi}}\times \sqrt 3 $$
or, $$h=\sqrt{\cfrac{3A}{\pi}}$$
Question 7
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A tower subtends an angle $$\alpha$$ at a point on the same level as the root of the tower and at a second point, $$b$$ meters above the first, the angle of depression of the foot of the tower is $$\beta$$. The height of the tower is

A
$$b\cot { \alpha } \tan { \beta } $$

B
$$b\tan { \alpha } \tan { \beta } $$

C
$$b\tan { \alpha } \cot { \beta } $$

D
None of these

Solution

From the figure, in $$\triangle ABD$$
$$\dfrac { AB }{ BD } =\tan { \alpha }$$
$$\Rightarrow \dfrac { h }{ x } =\tan { \alpha }$$
$$\Rightarrow h=x\tan { \alpha } \quad\quad\quad\dots(i)$$

In right angled $$\triangle BCE$$
$$\dfrac { BE }{ EC } =\tan { \beta }$$
$$\Rightarrow \dfrac { b }{ x } =\tan { \beta }$$
$$\Rightarrow x=\dfrac { b }{ \tan { \beta } } \quad\quad\quad\dots(ii)$$

Now, from equation $$(i)$$ and $$(ii),$$
$$h=\dfrac { b }{ \tan { \beta } } \times \tan { \alpha }$$
$$=b\tan { \alpha } \cot { \beta } $$

Hence, height of the tower is equal to $$b\tan\alpha\cot\beta.$$
Question 8
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A flag is mounted on the semi-circular dome of radius $$r$$. The elevation of the top of the flag at some point on the ground is $$30^{0}$$. Moving $$d$$ distance towards the dome, when the flag is just visible the angle of elevation is $$45^{0}$$, then the relation between $$r$$ and $$d$$ is

A
$$r=\displaystyle \frac{d}{\sqrt{2}(\sqrt{3}-1)}$$

B
$$r=\displaystyle \frac{d(2\sqrt{2})}{\sqrt{3}+1}$$

C
$$r=\displaystyle \frac{d}{\sqrt{2}(\sqrt{3}+1)}$$

D
$$r=\displaystyle \frac{d(2\sqrt{2})}{\sqrt{3}-1}$$

Solution

Let $$PQ=h=$$height of flag staff
r=radius of hemi-spherical dome
Elevation of top Q of a flag staff as seen from A is $$30°$$
$$\therefore r+h=OA\tan { 30° } \rightarrow 1$$
On walking distance 'd', the top is just visible from B
$$\therefore BQ$$ is tangent at R
$$\angle OBQ=45°=\angle OQB$$
$$\therefore OB=OQ=r+h$$
& $$OA=r+h+d$$
$$\therefore r=OR=RQ=RB$$
& $$BQ=BR+PQ=2r$$
In $$\triangle OBQ$$
$$\Rightarrow h=r\left( \sqrt { 2 } -1 \right) \rightarrow 2$$
From $$1$$, $$r+h=\left( r+h+d \right) \cfrac { 1 }{ \sqrt { 3 } } $$
$$\left( \sqrt { 3 } -1 \right) \sqrt { 2 } r=d$$
$$\Rightarrow r=\cfrac { d }{ \sqrt { 2 } \left( \sqrt { 3 } -1 \right) } $$
Question 9
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The angle of elevation of the summit of a mountain at a point $$45^{0}$$. After walking $$200$$ mt from $$A$$ towards the mountain along a road inclined at $$15^{0}$$, it is observed that the angle of elevation of the summit is $$60^{0}$$. The height of the mountain is

A
$$100(\sqrt{6}+\sqrt{2})$$ mt

B
$$100(\sqrt{6}-\sqrt{2})$$ mt

C
$$\displaystyle \frac{100}{\sqrt{6}}$$ mt

D
$$\displaystyle \frac{100}{\sqrt{2}}$$ mt

Solution

In $$\triangle ABC$$
$$\tan 45^o=\dfrac{H}{BC}\Rightarrow BC=H$$

We, know that,
$$DB=200$$

Therefore, $$y=200\sin 15^o\\x=200\cos 15^o$$

Now, $$AE=H-200\sin 15^o\\DE=H-200\cos 15^o$$

in $$\triangle ADE$$
$$\tan 60^o=\dfrac{AE}{DE}\\ \Rightarrow \sqrt3=\dfrac{H-200\sin 15^o}{H-200\cos 15^o} \\ \Rightarrow \sqrt3=\dfrac{H-200(\dfrac{\sqrt3-1}{2\sqrt2})}{H-200(\dfrac{\sqrt3+1}{2\sqrt2})}$$
$$[2\sqrt2 H-200(\sqrt{3}+1)]\sqrt{3} =H2\sqrt2-200(\sqrt{3}-1)$$
$$2\sqrt {2} H (\sqrt3 -1)=200(3+\sqrt 3- \sqrt 3+1)$$

$$H=\dfrac{800}{2\sqrt2(\sqrt3-1)}$$

$$=\dfrac{400 \times \sqrt 2 (\sqrt 3+1)}{\sqrt2(\sqrt 3 -1) \times \sqrt 2 (\sqrt 3+1)}$$ (Rationalising the denominator )

$$=\dfrac{400 (\sqrt6+\sqrt 2)}{4}$$

$$=100(\sqrt6+\sqrt2)$$

Therefore, height of the mountain is $$100(\sqrt6+\sqrt2)$$
Question 10
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The angular depression of the top and the foot of a tower as seen from the top of a second tower which is $$150m$$ high and standing on the same level as the first are $$\alpha$$ and $$\beta$$ respectively. If $$\tan\alpha=\dfrac{4}{3}$$ and $$\tan\beta=\dfrac{5}{2}$$, the distance between their tops is:

A
$$100m$$

B
$$120m$$

C
$$110m$$

D
$$130m$$

Solution

Let $$AB$$ be the length of $$1^{st}$$ tower.

Let $$BC$$ be the length of $$2^{nd}$$ tower.

Here given that CD$$=150 m$$

$$BE=CD=150m$$

From $$\triangle BCE$$

$$\tan \beta=\cfrac{BE}{CE}$$

or, $$\cfrac{5}{2}=\cfrac{150}{CE}$$

or, $$CE=150\times \cfrac{2}{5}=60$$

$$\tan\alpha=\cfrac{4}{3}$$

$$\sec\alpha=\sqrt {1+\cfrac{16}{9}}\\\ \ \ \ \ \ \ \ =\sqrt {\cfrac{25}{9}}\\\ \ \ \ \ \ \ =\cfrac{5}{3}$$[only positive value taken].

Thus,

$$\cos\alpha=\cfrac{3}{5}$$

$$\cos\alpha=\cfrac{CE}{AC}$$

or, $$\cfrac{60}{AC}=\cfrac{3}{5}$$

or, $$AC=\cfrac{5}{3} \times 60=100$$

$$\therefore$$ Distance between their tops$$=100m$$