Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 16

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Question 1
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The upper part of a tree broken over by the wind makes an angle of $$60^{0}$$ with the ground and touches the ground at a distance of 50 metres from the foot. The height of the tree in metres is

A
$$124.2$$

B
$$186.6$$

C
$$243.2$$

D
$$164.2$$

Solution

We will consider that the tree is broken at A.
So the total heigh of the tree will be $$AC + AB$$

Let $$AC=L_1$$ and $$AB =L_2$$
$$\Rightarrow$$Height of the tree $$= L_{1} + L_{2}$$
In the $$\triangle ABC, \angle C = 90^\circ$$,

$$\sin 60^\circ = \dfrac{AC}{AB}$$
$$L_{2} \sin 60^{0} = L_{1}$$ .....(1)
$$\tan 60^{0} = \displaystyle\frac{L_{1}}{50}$$
$$\Rightarrow L_{1} = 50 \sqrt{3} m$$
Substituting the value of $$L_1$$ in (1)
$$L_2 \times \dfrac{\sqrt 3}{2} = 50\sqrt3$$
$$\Rightarrow L_{2} = 100 m$$
$$\Rightarrow h =L_{1}+L_{2}$$
$$\Rightarrow h = 100+50 \sqrt{3}$$
$$=186.6 m$$
$$\therefore$$ Height of the tree = $$=186.6 m$$
Question 2
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The angle of elevation of the top of a flagstaff when observed from a point at a distance $$60$$ meters from its foot is $$30^{0}$$. The height of the flagstaff (in meters) is:

A
$$20\sqrt{3}$$

B
$$10\sqrt{3}$$

C
$$60\sqrt{3}$$

D
$$30\sqrt{3}$$

Solution

Let $$AB=h$$ be the height of the flagstaff and $$C$$ be the point at a distance of $$60m$$ from the foot.
In $$\triangle ABC$$,

$$\tan{30^0} = \dfrac{AB}{BC} = \dfrac{h}{60}$$

$$\Rightarrow \dfrac1{\sqrt3} = \dfrac h{60}$$

$$\Rightarrow h = \dfrac{60}{\sqrt3}$$

$$\Rightarrow h = 20\sqrt3m$$
Question 3
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From the top of a hill $$h$$ meters high, the angle of depression of the top and the bottom of a pillar are $$\alpha,\ \beta$$ respectively. Then the height(in meters) of the pillar is

A
$$\displaystyle \dfrac{h(\tan\beta-\tan\alpha)}{\tan\beta}$$

B
$$\displaystyle \dfrac{h(\tan\alpha-\tan\beta)}{\tan\alpha}$$

C
$$\displaystyle \dfrac{h(\tan\beta+\tan\alpha)}{\tan\beta}$$

D
$$\displaystyle \dfrac{h(\tan\beta+\tan\alpha)}{\tan\alpha}$$

Solution

$$\tan \alpha = \dfrac{h-d}{a}$$
$$\tan \beta = \dfrac{h}{a}$$

$$\dfrac{\tan \alpha}{\tan \beta} = \dfrac{h-d}{h}$$

$$\Rightarrow \dfrac{\tan \alpha}{\tan \beta} = 1- \dfrac{d}{h}$$

$$\Rightarrow \dfrac{d}{h}= \dfrac{\tan \beta-\tan \alpha}{\tan \beta}$$

$$\Rightarrow d= h\left(\dfrac{\tan \beta-\tan \alpha}{\tan \beta}\right)$$
Question 4
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From the top of the tree, a man observes the angle of depression of a point which is at a distance of $$40\ m$$ from the foot is $$75^{0}$$. The height of the tree is:

A
$$40\sqrt{3}\ m$$

B
$$40(2+\sqrt{3})\ m$$

C
$$21\sqrt{3}\ m$$

D
$$3\sqrt{21}\ m$$

Solution

Let say AC be the tree and the Observer is at A who is observing point B.
Given: BC = 40m
$$\angle B = 75^o $$ = angle of depression {alternate angles}

In $$\triangle ABC$$,

$$\tan 75^0 = \dfrac{h}{40}$$

$$\Rightarrow \tan (45^o+30^o) = \dfrac{h}{40}$$

$$\Rightarrow \dfrac{\tan45^o+ \tan 30^o}{1-\tan45^o\tan 30^o} = \dfrac{h}{40}$$ $$\because \dfrac{\tan A+ \tan B}{1-\tan A\tan B} = tan(A+B)$$

$$\Rightarrow \dfrac{1+\dfrac{1}{\sqrt{3}}}{1- 1\times \dfrac{1}{\sqrt{3}}}= \dfrac{h}{40}$$

$$\Rightarrow \dfrac{(\sqrt{3}+1)}{(\sqrt{3}- 1)} = \dfrac{h}{40}$$

$$\Rightarrow\left(\dfrac{3+1+2\sqrt{3}}{2}\right) = \dfrac{h}{40}$$ {rationalizing by multiplying and dividing by $$\sqrt 3+1$$}

$$\Rightarrow (2+\sqrt{3}) = \dfrac{h}{40}$$

$$\Rightarrow h = 40 (2+\sqrt{3})m$$
Question 5
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A kite is flying with the string inclined at $$30^{o}$$ to the horizon. The height of the kite above the ground, when the string is $$15\ m$$ long is

A
$$15\ m$$

B
$$30\ m$$

C
$$\dfrac{15}2\ m$$

D
$$\dfrac{15}3\ m$$

Solution

$$\sin{30^0} = \dfrac{AB}{AC} = \dfrac h{15}$$

$$\Rightarrow \dfrac12 = \dfrac h{15}$$

$$\Rightarrow h = \dfrac {15}2\ m$$
Question 6
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A pole stands vertically, inside a triangular park $$ABC$$. If the angle of elevation of the top of the pole from each corner of the park is same, then the foot of the pole is at the:

A
centriod

B
circumcentre

C
incentre

D
orthocentre

Solution

If angle of elevation$$(p)$$ is same from all the vertices then the Pole must be at Equal Distance from each of the vertices which will be $$=R\ (Circumradius)$$

as $$\tan p=\dfrac{H}{R}\Rightarrow R=\dfrac{H}{\tan p}$$[which is same from all vertices]

therefore, the pole must be lying on the $$Circumcenter$$
Question 7
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The angles of elevation of top of a pole from two points $$A$$and $$B$$ on the horizontal line lying on opposite side of the pole are observed to be $$30^{\circ}$$ and $$60^{\circ}$$ If $$AB=100\ m$$, then height of the pole is

A
$$20\sqrt{3}\ m$$

B
$$15\sqrt{3}\ m$$

C
$$10\sqrt{3}\ m$$

D
$$25\sqrt{3}\ m$$

Solution

Let the height of pole be $$PQ=h$$ m.
And $$QB=d$$ be the distance between $$B$$ and foot of pole.
$$\tan 60 ^{\circ} = \dfrac{h}{d}$$
$$\sqrt3=\dfrac{h}{d}$$

$$\Rightarrow d = \dfrac{h}{\sqrt{3}}$$....(1)

$$\tan 30 ^{\circ} = \dfrac{h}{100-d}$$

$$\dfrac{1}{\sqrt3} = \dfrac{h}{100-d}$$
$$\Rightarrow 100-d={\sqrt3}h$$
Substitute the value of d from the equation (1), we get
$$\Rightarrow 100- \dfrac{h}{\sqrt{3}} = \sqrt{3} h$$

$$\Rightarrow 100 =h\left(\sqrt3+ \dfrac{1}{\sqrt{3}}\right)$$

$$\Rightarrow 100=h ({\dfrac{3+1}{\sqrt3}})$$

By doing cross multiplication
$$\Rightarrow 100{\sqrt3}=4h$$
$$\Rightarrow 4h=100{\sqrt3}$$
$$\Rightarrow h = 25 \sqrt{3}\ m$$
Question 8
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lf the shadow of a tower is $$\sqrt{3}$$ times of its height, the altitude of the sun is

A
$$15^{0}$$

B
$$30^{0}$$

C
$$45^{0}$$

D
$$60^{0}$$

Solution

Let the height of tower be $$h$$.
$$\tan \theta = \dfrac{h}{\sqrt{3}h}$$

$$\Rightarrow \tan \theta = \dfrac{1}{\sqrt{3}}$$

$$\Rightarrow \theta = 30^{o}$$
Question 9
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The shadow of a tower on a level plane is found to be $$60$$ metres longer when the sun's altitude is $$30^{0}$$ than that when it is $$45^{0 }$$. The height of the tower in metres is

A
$$30(\sqrt{3}+1)$$

B
$$30(\sqrt{3}-1)$$

C
$$30(3+\sqrt{3})$$

D
$$30(3-\sqrt{3})$$

Solution

Let height of tower be $$h \ m$$.
And length of shadow when sun's altitude is at $$30^0$$.
$$\tan 45^0 = \dfrac{h}{L_{1}}$$
$$\Rightarrow L_{1} = h$$

$$\tan 30^0 = \dfrac{h}{L_{1}+60}$$

$$\Rightarrow \dfrac{1}{\sqrt{3}}= \dfrac{L_{1}}{L_{1} + 60}$$

$$\Rightarrow \sqrt{3} = 1+ \dfrac{60}{L_{1}}$$

$$\Rightarrow L_{1} = \dfrac{60}{(\sqrt{3}-1)}$$

$$\Rightarrow L_{1} = 30 (\sqrt{3}+1)\ m$$
Question 10
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$$A$$ man on the top of an observation tower finds an object at an angle of depression $$30^{0}$$. After the object was moved $$30$$ metres in a straight line towards the tower, he finds the angle of depression to be $$45^{0}$$. The distance of the object now from the foot of the tower in metres is

A
$$15\sqrt{3}$$

B
$$15(\sqrt{3}+1)$$

C
$$15(\sqrt{3}-1)$$

D
$$15(2+\sqrt{3})$$

Solution

Let the height of the tower be $$h$$
$$\implies AD=h$$
And $$d$$ be the initial distance of object from tower.
$$\implies BD=d$$
So, $$CD=d-30$$
In $$\triangle ACD$$
$$\tan 45^0 = \dfrac{h}{d -30 }$$
$$\implies 1=\dfrac{h}{d-30}$$
$$\implies h=d-30$$
In $$\triangle ABD$$
$$\tan 30^0 = \dfrac{h}{d}$$
$$\implies \dfrac{1}{\sqrt3}=\dfrac{h}{d}$$
$$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{d-30}{d}$$ (on putting $$h=d-30$$)

$$\Rightarrow \dfrac{30}{d} = 1- \dfrac{1}{\sqrt{3}}$$

$$\Rightarrow d = \dfrac{30 \times \sqrt{3}}{(\sqrt{3}-1)}$$

Now, the distance $$= d-30$$
$$= \dfrac{30\sqrt{3}}{(\sqrt{3}-1)} -30$$
$$= \dfrac{30}{(\sqrt{3}-1)}$$
$$= 15(\sqrt{3}+1)\ m$$