Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 57

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Question 1
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A vertical tower $$CP$$ subtends the same angle $$\theta$$, at point $$B$$ on the horizontal plane through $$C$$, the foot of the tower, and at point $$A$$ in the vertical plane. If the triangle $$ABC$$ is equilateral with length of each side equal to $$4m$$, then height of the tower is:

A
$$8\sqrt{3}m$$

B
$$\displaystyle \frac{4\sqrt{3}}{3}m$$

C
$$4\sqrt{3}m$$

D
$$\displaystyle \frac{8}{\sqrt{3}}m$$
Question 2
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$$n$$ poles standing at equal distances on a straight road subtend the same angle $$\alpha$$ at a point $$O$$ on the road. If the height of the largest pole is $$h$$ and the distance of the foot of the smallest pole from $$O$$ is $$a$$, the distance between two consecutive poles is:

A
$${\dfrac{h\sin\alpha-a\cos\alpha}{(n-1)\sin\alpha}}$$

B
$$\displaystyle \dfrac{h\cos\alpha-a\cos\alpha}{(n-1)\cos\alpha}$$

C
$$\displaystyle \dfrac{h\cos\alpha-a\sin\alpha}{(n-1)\sin\alpha}$$

D
$$\displaystyle \dfrac{h\sin\alpha-a\cos\alpha}{(n-1)\cos\alpha}$$

Solution

$$\cfrac{h}{a+(n-1)x}=\cfrac{y}{a}\cdot y=a\tan \alpha$$
or, $$\cfrac{h}{a+(n-1)x}=\cfrac{y}{a}\cdot y=\cfrac{a\tan\alpha}{a}$$
or, $$\cfrac{h}{a+(n-1)x}=\cfrac{y}{a}\cdot y=\tan\alpha$$
or, $$\cfrac{h}{\tan\alpha}=a+(n-1)x$$
or, $$(n-1)x=h-a\tan\alpha$$
or, $$x=\cfrac{h-a\tan\alpha}{(n-1)\times \tan\alpha}$$
or,$$x=\cfrac{h\cos\alpha-a\sin\alpha}{(n-1)\cos\alpha\times \tan\alpha}$$
or, $$x=\cfrac{h\cos\alpha-a\sin\alpha}{(n-1)\sin\alpha}$$
Question 3
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A $$6$$ ft-tall man finds that the angle of elevation of the top of a $$24$$ ft-high pillar and the angle of depression of its base are complementary angles.The distance of the man from the pillar is

A
$$\displaystyle 2\sqrt{3}\ ft$$

B
$$\displaystyle 8\sqrt{3}\ ft$$

C
$$\displaystyle 6\sqrt{3}\ ft$$

D
none of these

Solution

Let the distance between the tower and the man be x ft
$$\Rightarrow AP=CQ=x\ ft$$
In rt $$\Delta BCQ$$
$$\tan \alpha=\displaystyle \dfrac{18}{x}$$ ....(1)

In rt $$\Delta ACQ$$
$$\tan (90-\alpha)=\displaystyle \dfrac{6}{x}$$
$$\Rightarrow \cot \alpha=\displaystyle \dfrac{6}{x}$$ ....(2)

Multiplying (1) and (2), we get
$$\Rightarrow tan(\alpha)\times cot(\alpha)=\dfrac{18\times 6}{x\times x}$$
$$\Rightarrow 1=\dfrac{18\times 6}{x^2}$$
$$\Rightarrow x^2=36\times 3$$
$$\Rightarrow x=\sqrt{36\times 3}$$
$$\Rightarrow x=6\sqrt{3}\ ft$$
Question 4
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A piece of paper in the shape of a sector of a circle of radius 10cm and of angle $$\displaystyle 216^{\circ}$$ just covers the lateral surface of a right circular cone of vertical angle$$\displaystyle 2\theta$$ .Then $$\displaystyle \sin\: \theta$$ is

A
$$\displaystyle \frac{3}{5}$$

B
$$\displaystyle \frac{4}{5}$$

C
$$\displaystyle \frac{3}{4}$$

D
none of these
Question 5
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At the foot of the mountain the elevation of its summit is $${ 45 }^{ 0 }$$; after ascending $$1000m$$ towards the mountain up a slope of $${ 30 }^{ 0 }$$ inclination, the elevation is found to be $${ 60 }^{ 0 }$$. The height of the mountain is

A
$$\displaystyle \frac { \sqrt { 3 } +1 }{ 2 } m$$

B
$$\displaystyle \frac { \sqrt { 3 } -1 }{ 2 } m$$

C
$$\displaystyle \frac { \sqrt { 3 } +1 }{ 2\sqrt { 3 } } $$

D
none of these

Solution

Let $$P$$ be the summit of the mountain and $$Q$$ be the foot.

Let $$A$$ be the first position and $$B$$ the second position of observation.

$$BN$$ and $$BM$$ are $$\bot$$s to $$PQ$$ and $$AQ$$ respectively.

Then $$AB=1000m=1km$$,

$$\angle MAB={ 30 }^{ 0 },\angle MAP={ 45 }^{ 0 },\angle NBP={ 60 }^{ 0 }$$

Now, $$\angle BAP=\angle MAP-\angle MAB={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$$

$$\angle APB=\angle APN-\angle BPN={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$$ .........$$[ \angle BNP = 90^{\circ}, \angle BPN = 90- 60=30^{\circ}]$$

$$\therefore \triangle ABP$$ is isosceles and $$\therefore AB=BP$$

But $$AB=1$$ kilometer, $$\therefore BP=1$$ kilometer

Now $$PQ=PN+NQ=PN+BM$$

$$=BP\sin { { 60 }^{ 0 } } +AB\sin { { 30 }^{ 0 } } $$ ......... $$ \because sin(\Theta ) = \dfrac {Opposite}{Hypotenuse} $$ for right $$\angle \triangle$$

$$\displaystyle =1.\frac { \sqrt { 3 } }{ 2 } +1.\frac { 1 }{ 2 } =\frac { \sqrt { 3 } +1 }{ 2 } m$$

The height of the mountain is $$\dfrac { \sqrt { 3 } +1 }{ 2 } m$$.
Question 6
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A man from the top of a $$100$$-metre-high tower sees a car moving towards the tower at an angle of depression of $$\displaystyle 30^{\circ}$$. After some time, the angle of depression becomes $$\displaystyle 60^{\circ}$$. The distance (in metres) travelled by the car during the time, is

A
$$\displaystyle 100\sqrt{3}$$

B
$$\displaystyle \frac{200\sqrt{3}}{3}$$

C
$$\displaystyle \frac{100\sqrt{3}}{3}$$

D
$$\displaystyle 200\sqrt{3}$$

Solution

In $$\Delta ABD$$
$$\tan 60^{0}=\displaystyle \dfrac{100}{x}$$
$$\Rightarrow x=\displaystyle \dfrac{100\sqrt{3}}{3} m$$

In $$\Delta ABC$$
$$\tan 30^{0}=\displaystyle \dfrac{100}{x+y}$$

$$\Rightarrow \displaystyle \dfrac{1}{\sqrt{3}}=\dfrac{100\times3}{100\sqrt{3}+3y}$$

$$\Rightarrow y=\displaystyle \dfrac{200\sqrt{3}}{3} m$$
Question 7
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A vertical lamp-post, 6 m high, stands at a distance of 2 m from a wall, 4 m high. A 1.5-m-tall man starts to walk away from the wall on the other side of the wall, in line with the lamp-post.The maximum distance to which the man can walk remaining in the shadow is

A
$$\displaystyle \displaystyle \frac{5}{2}m$$

B
$$\displaystyle \frac{3}{2}m$$

C
$$\displaystyle 4m$$

D
none of these

Solution

From similar $$\triangle ABQ$$ and $$\triangle CDQ$$, we get
$$\cfrac { BR }{ DR } =\cfrac { AB }{ CD } =\cfrac { 4 }{ \cfrac { 3 }{ 2 } } =\cfrac { 8 }{ 3 } $$ ...(1)
Now, from similar $$\triangle PQR$$ and $$\triangle ABR$$, we get
$$\cfrac { PQ }{ AB } =\cfrac { QR }{ BR } \Rightarrow \cfrac { 6 }{ 4 } =\cfrac { BQ+BR }{ BR } =\cfrac { 2+BR }{ BR } $$
$$\Rightarrow BR=4$$ ...(2)
Substituting (2) in (1), we get
$$\cfrac { 4 }{ DR } =\cfrac { 8 }{ 3 } \Rightarrow DR=\cfrac { 3 }{ 2 } $$ ...(3)
From (2) and (3)
$$BD=BR-DR=4-\cfrac { 3 }{ 2 } =\cfrac { 5 }{ 2 } $$
Question 8
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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 circumscribing the hexagon be $$A\ \displaystyle m^{2},$$ then the area of the hexagon is

A
$$\displaystyle \frac{3\sqrt{3}}{8}A\: m^{2}$$

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

C
$$\displaystyle \frac{3\sqrt{3}}{4\pi }A\: m^{2}$$

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

Solution

Let a be the side of the regular hexagon
Now, for the equilateral triangle FOA, we have
$$OF=OA=AF=a$$
Hence area of the circle $$\displaystyle A=\pi { a }^{ 2 }\Rightarrow a=\sqrt { \frac { A }{ \pi } } $$
Now, area of the hexagon $$= \dfrac{3\sqrt3}2 a^2$$

$$=\dfrac{3\sqrt3}{2\pi}A\ m^2$$
Question 9
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$$AB$$ is a vertical pole with base at $$B$$. A man finds the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $$60^{0}$$ . He move away from the pole along line $$BC$$ to a point $$D$$ such that $$CD= 7\:m$$ , from $$D$$ the angle of elevation of point $$A$$ is $$45^{0}$$ , then height of the pole is

A
$$\displaystyle \frac{7\sqrt{3}}{2}\left ( \sqrt{3}+1 \right )$$

B
$$\displaystyle \frac{7\sqrt{3}}{2}\left ( \sqrt{3}-1 \right )$$

C
$$\displaystyle \frac{7\sqrt{3}}{2\left ( \sqrt{3}+1 \right )}$$

D
$$\displaystyle \frac{7\sqrt{3}}{2\left ( \sqrt{3}-1 \right )}$$

Solution

Let $$AB=x$$ and $$BC=y$$
In $$\triangle ABC\quad \tan 60=\cfrac { x }{ y } $$
$$\Rightarrow y=\cfrac { x }{ \sqrt { 3 } } $$ ..(1)
And in $$\triangle ABD\quad \tan 45=\cfrac { x }{ y+7 } $$
From (1)
$$ x=y+7\\=\cfrac { x }{ \sqrt { 3 } } +7$$
$$\sqrt3x-x=7\sqrt3$$
$$x=\dfrac{7\sqrt3}{\sqrt3-1}$$
$$\Rightarrow x=\cfrac { 7\sqrt { 3 } }{ 2 } \left( \sqrt { 3 } +1 \right) $$
Question 10
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At the foot of the mountain the elevation of its summit is $${45}^{0}$$; after ascending $$1000\ m$$ towards the mountain up a slope of $${30}^{0}$$ inclination, the elevation is found to be $${60}^{0}$$. The height of the mountain is

A
$$\displaystyle \frac { \sqrt { 3 } +1 }{ 2 } $$ $$km$$

B
$$\displaystyle \frac { \sqrt { 3 } -1 }{ 2 } $$ $$km$$

C
$$\displaystyle \frac { \sqrt { 3 } +1 }{ 2\sqrt { 3 } } $$ $$km$$

D
None of these

Solution

Let $$P$$ be the summit of the mountain and $$Q$$ be the foot.
Let $$A$$ be the first position and $$B$$ the second position of observation.
$$BN$$ and $$BM$$ are perpendiculars from $$B$$ to $$PQ$$ and $$AQ$$ respectively.
Then $$AB=1000 m=1 km$$
$$\angle MAB={ 30 }^{ 0 };\angle MAP={ 45 }^{ 0 };\angle NBP={ 60 }^{ 0 }$$
Now, $$\angle BAP=\angle MAP-\angle MAB={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$$
$$\angle APB=\angle APN-\angle BPN={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$$
$$\therefore \triangle ABP$$ is isosceles and $$\therefore AB=BP$$
But $$AB=1$$ kilometer, $$\therefore BP=1$$ kilometer
Now, $$PQ=PN+NQ=PN+BM$$
$$=BP\sin { { 60 }^{ 0 } } +AB\sin { { 30 }^{ 0 } } $$
$$\displaystyle =1.\frac { \sqrt { 3 } }{ 2 } +1.\frac { 1 }{ 2 } =\frac { \sqrt { 3 } +1 }{ 2 } $$ $$km$$