Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 29

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Question 1
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The angular elevation of a tower from a point is $$\displaystyle 30^{\circ}$$ As we more $$100$$ m nearer to the base of the tower the angle of elevation becomes $$\displaystyle 60^{\circ}$$. Find the height of the tower and the distance of the first point from the base of the tower

A
$$150 m$$

B
$$90 m$$

C
$$120 m$$

D
$$180 m$$

Solution

Let AB represent the tower $$P, Q$$ the points of observation It is given $$PQ = 100$$ m Angles of elevation at $$P$$ and $$Q$$ are $$\displaystyle 30^{\circ}$$ and $$\displaystyle 60^{\circ}$$ respectively
Let the required height of the tower $$AB = h$$ metres and $$BQ - x$$ metres Then in $$\displaystyle \Delta BAQ,\dfrac{h}{x}=\tan 60^{\circ}$$
$$\displaystyle \Rightarrow h=x\tan 60^{\circ}=x\sqrt{3}..........(i)$$
In $$\displaystyle \Delta ABP,\dfrac{h}{10+x}=\tan 30^{\circ}$$
$$\displaystyle \Rightarrow \dfrac{h}{100+x}=\dfrac{1}{\sqrt{3}}\Rightarrow h\sqrt{3}=100+x$$
$$\displaystyle \Rightarrow x\sqrt{3}\times \sqrt{3}=100+x\Rightarrow 3x=100\Rightarrow 2x=100\Rightarrow x=50$$
$$\displaystyle \therefore$$ From (i) $$\displaystyle h=50\sqrt{3m}=50\times 1.732m=86.6m$$
Also $$AP = (x + 100)$$ m = $$(50 + 100)$$ m = $$150$$ m
Question 2
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The angle of elevation of the top of a tower from point at a distance of $$100$$ metres from its foot on a horizontal plane is found to be $$\displaystyle 60^{\circ}$$. Find the height of the tower

A
$$173.2 m$$

B
$$196 m$$

C
$$144 m$$

D
$$123.2 m$$

Solution

Let the height of the tower AC be h metres Given distance BC = 100 m, $$\displaystyle \angle ABC=60^{\circ}$$
$$\displaystyle \therefore \tan 60^{\circ}=\frac{AC}{BC}=\frac{h}{100}\Rightarrow h=100\times \tan 60^{\circ}=(100\times \sqrt{3})metre$$
$$\displaystyle =100\times 1.732=173.2\: m$$
Question 3
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The length of a shadow of a pole is $$\displaystyle \sqrt{3} $$ times the length of the pole. The angle of elevation of the sun is

A
$$\displaystyle 30^{\circ} $$

B
$$\displaystyle 60^{\circ} $$

C
$$\displaystyle 90^{\circ} $$

D
$$\displaystyle 45^{\circ} $$

Solution

Length of shadow (s) = $$\sqrt{3}$$ Length of the pole (p)
The shadow and a pole will form a right angled triangle, with $$\theta$$ being the angle of elevation
Hence, $$\tan \theta = \dfrac{p}{s}$$
$$\tan \theta = \dfrac{p}{p \sqrt{3}}$$
$$\tan \theta = \tan 30$$
$$\theta = 30^{\circ}$$
Question 4
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Two men on either side of a temple $$126\ m$$ high observe the angle of elevation of the top of the temple to be $$\displaystyle 30^{\circ}$$ and $$\displaystyle 60^{\circ}$$ respectively. Find the distance between the two men?

A
$$\displaystyle 168\sqrt{3}\ m$$

B
$$\displaystyle 180\sqrt{3}\ m$$

C
$$\displaystyle 160\sqrt{3}\ m$$

D
$$\displaystyle 148\sqrt{3}\ m$$

Solution

In $$\Delta ABC,$$
$$\tan 30^{\circ}=\dfrac{AB}{BC}$$

$$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{126}{x}$$

$$\Rightarrow x=126\sqrt{3}$$

Now, in $$\Delta ABD,$$
$$\tan 60^{\circ}=\dfrac{AB}{BD}$$

$$\Rightarrow \sqrt{3}=\dfrac{126}{y}$$

$$\Rightarrow y=\dfrac{126}{\sqrt{3}}$$

$$\Rightarrow y=\dfrac{126\sqrt{3}}{3}\ m$$

$$\Rightarrow y = 42\sqrt{3}\ m$$

$$\therefore$$ Required distance $$= x + y$$
$$=126\sqrt{3}\ m+42\sqrt{3}\ m$$
$$=168\sqrt{3}\ m$$
Question 5
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Two pillars are of equal height on either sides of a road which is 100 m wide. The angles of elevation of the top of the pillars are $$\displaystyle 60^{\circ}$$ and $$\displaystyle 30^{\circ}$$ at a point on the road between the pillars. Find the position of the point between the pillars and height of each pillar

A
$$43.3 m$$

B
$$28 m$$

C
$$45.3 m$$

D
$$54 m$$

Solution

Let AB and ED be two pillars each of height h metres Let C be a point on he road BD such that
BC = x metres Then CD = (100 - x) metres Given $$\displaystyle \angle ACB=60^{\circ}$$ and $$\displaystyle \angle ECD=30^{\circ}$$
In $$\displaystyle \Delta ABC, \tan 60^{\circ}=\frac{AB}{BC}$$ $$\displaystyle \Rightarrow \sqrt{3}=\frac{h}{x}\Rightarrow h=\sqrt{3}x..........(i)$$
In $$\displaystyle \Delta ECD, \tan 30^{\circ}=\frac{ED}{CD}$$ $$\displaystyle \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{100-x}\Rightarrow h\sqrt{3}=100-x...........(ii)$$
$$\therefore$$ Subst. the value of h from (i) in (ii) we get
$$\displaystyle \sqrt{3}.x=\frac{100-x}{\sqrt{3}}\Rightarrow 3x=100-x\Rightarrow 4x=100\Rightarrow x=25m$$
$$\therefore$$ $$\displaystyle h=(\sqrt{3}\times 25)=25\times 1.732m=43.3m$$
$$\therefore$$ The required point is at a distance of 25 m from the pillar B and the height of each pillar is 43.3 m
Question 6
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A circus artist is climbing a 20 m long rope Which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is $$\displaystyle 30^{\circ}$$

A
$$10 \ m$$

B
$$9 \ m$$

C
$$12 \ m$$

D
$$8\ m$$

Solution

Based on the given information, we can draw the figure shown above.
Let the height of the pole be $$h\ m$$.
and $$AC = 20 \ m$$ be the rope which the circus artist is climbing.
Given, $$\displaystyle \angle ACB=30^{\circ}$$

We know that, $$\sin \theta=\dfrac{\text{Opposite Side}}{\text{Hypotenuse}}$$

Hence, in $$\displaystyle \Delta ABC$$,
$$\displaystyle \sin 30^{\circ}=\displaystyle \dfrac{AB}{AC}\\$$
$$=\dfrac{h}{20}\\$$
$$\displaystyle \Rightarrow \dfrac{h}{20}=\frac{1}{2}\quad \quad \left[\because \ \sin 30^o=\dfrac12\right]\\$$
$$\Rightarrow h=\dfrac{1}{2}\times 20$$
$$=10\ m$$

Hence, the height of the pole is $$10\ m$$.
Question 7
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A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. If the length of the string is $$\displaystyle 40\sqrt{3}\: m$$ then find the inclination of the string with the ground.

A
$$\displaystyle \theta =60^{\circ}$$

B
$$\displaystyle \theta =30^{\circ}$$

C
$$\displaystyle \theta =45^{\circ}$$

D
$$\displaystyle \theta =15^{\circ}$$

Solution

Let $$\displaystyle \theta$$ be the inclination of the kite with the ground.
Height of kite above horizontal $$= KP = 60 m$$
Length of string of kite $$= OK =$$ $$\displaystyle 40\sqrt{3}\: m$$
In $$\triangle KOP$$,
$$\displaystyle \therefore \sin \theta =\frac{KP}{OK}$$

$$=\dfrac{60}{40\sqrt{3}}$$

$$=\dfrac{3}{2\sqrt{3}}$$

$$=\dfrac{\sqrt{3}}{2}=\sin 60^{\circ}$$

$$\displaystyle \therefore \theta =60^{\circ}$$
Question 8
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The elevation of the top of hill at each of the three angular points X, Y, Z of a horizontal $$\displaystyle \Delta $$ XYZ is $$\displaystyle \alpha $$ then the height of the hill is :

A
$$\displaystyle \frac{YZ\tan \alpha \cos \sec X}{2}$$

B
$$\displaystyle XZ \tan \alpha \cos \sec Y$$

C
$$\displaystyle \frac{XY\cot \alpha }{2\sin Z}$$

D
$$\displaystyle\frac{XY\tan \alpha }{2\sin Y}$$
Question 9
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The angle of elevation of the top of a tower as seen from two points $$A$$ & $$B$$ situated the same line and at distance '$$p$$' and '$$q$$' respectively from the foot of the tower are complementary, then height of the tower is

A
$$pq$$

B
$$\displaystyle \frac{p}{q}$$

C
$$\displaystyle \sqrt{pq}$$

D
none of these

Solution

Let $$OT$$ be the tower of height $$h$$
Let $$O$$ be the base of the tower
Let $$A$$ and $$B$$ be two points on the through the base such that $$OA = p$$ and $$OB = q$$
In $$\displaystyle \Delta AOT$$
$$\displaystyle \tan \alpha =\frac{OT}{OA}=\frac{h}{p}$$ .....(i)
In $$\displaystyle \Delta BOT$$
$$\displaystyle \Rightarrow \tan \left ( 90-\alpha \right )=\frac{OT}{OB}=\frac{h}{q}$$ or $$\cot \alpha =\dfrac{h}{q}$$ ......(ii)
Multiplying (i) and (ii), we have
$$\displaystyle \Rightarrow \tan \alpha \cot \alpha =\frac{h}{p}\times \frac{h}{q}$$
$$\displaystyle \Rightarrow \tan \alpha* \frac{1}{\tan \alpha }=\frac{h^{2}}{pq}$$
$$\displaystyle \Rightarrow 1=\frac{h^{2}}{pq}$$
$$\displaystyle \Rightarrow h^{2}=pq $$
$$\displaystyle \Rightarrow h=\sqrt{pq} $$
Question 10
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Vijay has been invited for dinner in a club. While walking through the garden path towards the building he observes that there is an electric rod on the top of the building.
From the point where he is standing the angles of elevation of the top of the electric rod and the top of the building are $$\displaystyle \Phi $$ and $$\theta$$ respectively.

If the heights of the electric rod and the building are $$p$$ and $$q$$ respectively, then mark all the correct statements

A
The height of the tower is $$\displaystyle \frac{p\tan\theta \tan \Phi }{\tan \Phi -\tan \theta } $$

B
The height of the electric rod is $$\displaystyle \frac{q\tan \theta}{\tan \Phi -\tan \theta}$$

C
The height of the tower is $$\displaystyle \frac{p\tan \theta }{\tan \Phi -\tan \theta}$$

D
The height of the electric rod is $$\displaystyle \frac{q\left ( \tan \Phi -\tan \theta \right )}{\tan \theta}$$

Solution

Based on the given information, we can draw the figure shown above.

We know that, $$\tan \theta=\dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$$

Hence, in $$\triangle ABC$$
$$\tan \phi=\dfrac{p+q}{x}$$

$$\Rightarrow x=\dfrac{p+q}{\tan \phi}..........(1)$$

Also, in $$\triangle DBC$$
$$\tan \theta=\dfrac{q}{x}$$

$$\Rightarrow x=\dfrac{q}{\tan \theta}..........(2)$$

Now, from $$(1)$$ and $$(2)$$, we get:
$$\dfrac{q}{\tan \theta}=\dfrac{p+q}{\tan \phi}$$

$$\Rightarrow q(\tan \phi)=(p+q)\tan \theta$$

$$\Rightarrow p\tan \theta=q(\tan \phi-\tan \theta)$$

$$\Rightarrow p=\dfrac{q(\tan \phi-\tan \theta)}{\tan \theta}$$

Hence, the height of electric rod is $$\dfrac{q(\tan \phi-\tan \theta)}{\tan \theta}$$.