Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 30

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Question 1
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A vertical pole stand at a point O on a horizontal ground. A and B are point on the ground 10m apart. The pole subtends an angle 30$$\displaystyle ^{\circ}$$ and 60$$\displaystyle ^{\circ}$$ at A and B, respectively, and AB subtends an angle 45$$\displaystyle ^{\circ}$$ at O. Then height of tower equals

A
$$\displaystyle 10\sqrt{3}$$

B
$$\displaystyle \frac{5}{2}\sqrt{3}$$

C
$$\displaystyle 5\sqrt{3}$$

D
None of these

Solution

Given that
$$\angle PAO=30^{\circ},\angle PBO=60^{\circ},AB=10\ m$$
Let $$PO$$ be the height of the tower
In $$\triangle PAO,\tan \angle PAO=\dfrac{PO}{OA}$$
$$\implies PO=OA\tan 30^{\circ}\implies OA=PO\sqrt{3}$$
In $$\triangle PBO,\tan \angle PBO=\dfrac{PO}{OB}$$
$$\implies PO=OB\tan 60^{\circ}\implies OB=\dfrac{PO}{\sqrt{3}}$$
$$ AB=10\implies OA-OB=10\implies PO\bigg(\sqrt{3}-\dfrac{1}{\sqrt{3}}\bigg)=10\implies PO=5\sqrt{3}$$
So height of the tower is $$5\sqrt{3}\ m$$
Question 2
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A man on a cliff observes a boat at an angle of depression of $$\displaystyle 30^{\circ} $$ which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later the angle of depression of the boat is found to be $$\displaystyle 60^{\circ}. $$ Find the total time taken by the boat from the initial point to reach the shore.

A
$$9$$ min

B
$$7$$ min

C
$$10$$ min

D
$$6$$ min
Question 3
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The angle of elevation of the top of tower as observed from a pint on the horizontal ground is 'x' if we move a distance 'd' towards the foot of the tower the angle of elevation increases to 'y' then the height of the tower is

A
$$\displaystyle \frac{d\tan x\tan y}{\tan y-\tan x}$$

B
$$\displaystyle d(\tan y+\tan x)$$

C
$$\displaystyle d(\tan y-\tan x)$$

D
$$\displaystyle \frac{d\tan x\tan y}{\tan y+\tan x}$$

Solution

Let the height of the tower be h.
$$\displaystyle \therefore $$ In $$\displaystyle \Delta ABC $$
$$\displaystyle \tan y=\dfrac{h}{b}$$
b = $$\displaystyle \dfrac{h}{\tan y}$$
In $$\displaystyle \Delta ABD $$
$$\displaystyle \tan x=\dfrac{h}{b+d}$$
$$\displaystyle \tan x=\dfrac{h}{\dfrac{h}{\tan y}+d}$$
$$\displaystyle \dfrac{h}{\tan x}=\dfrac{h}{\tan y}+d$$
$$d =$$ $$\displaystyle h\left ( \dfrac{1}{\tan x}-\dfrac{1}{\tan y} \right )$$
$$d =$$ $$\displaystyle h\left ( \dfrac{\tan y-\tan x}{\tan x\tan y} \right )$$
$$h =$$ $$\displaystyle \dfrac{d\tan x\tan y}{\tan y-\tan x}$$
Question 4
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The angle of elevation of the top of a tower as seen from two points $$A$$ & $$B$$ situated in the same line and at distances $$p$$ and $$q$$ respectively from the foot of the tower are complementary then the height of the tower is:

A
$$pq$$

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

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

D
None of these

Solution

Let the angle of elevation made at a distance of $$p =$$ $$\alpha$$
Then, angle of elevation made at a distance of $$q =$$ $$90 - \alpha$$
Let the height of tower $$=$$ $$h$$
Then, $$\tan \angle$$ of elevation $$=$$ $$\dfrac{Height}{distance}$$

Thus, $$\tan \alpha = \dfrac{h}{p}$$
$$\tan (90 - \alpha) = \dfrac{h}{q}$$ or $$\cot \alpha = \frac{h}{q}$$

Multiply both the equations,
$$\tan \alpha \cot \alpha = \dfrac{h}{p}. \dfrac{h}{q}$$
$$\rightarrow \dfrac{h^2}{pq} = 1$$
Or, $$h^2 = pq$$
$$h = \sqrt{pq}$$
Question 5
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The angle of elevation of the top of a tower at a distance of $$\displaystyle \frac{50\sqrt{3}}{3}$$ metres from the foot is $$\displaystyle 60^{\circ}$$. Find the height of the tower

A
$$\displaystyle 50\sqrt{3}$$ metres

B
$$\displaystyle \frac{20}{\sqrt{3}}$$ metres

C
$$-50$$ metres

D
$$50$$ metres

Solution

Let the height of the tower be $$h$$.
In $$ΔABC$$
$$\tan 60^{0}=\dfrac{AB}{BC}$$
$$=\dfrac { h }{ \dfrac { 50\sqrt { 3 } }{ 3 } } $$

$$=>\sqrt 3=\dfrac { 3\times h }{ { 50\sqrt { 3 } } } $$

$$=\dfrac { 3\times h }{ { 50\sqrt { 3 } \times \sqrt { 3 } } } $$

$$=\dfrac { 3\times h }{ { 50\times 3 } } $$

∴$$h=50$$m
Question 6
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If from the top of a tower 50 m high, the angles of depression of two objects due north of the tower are respectively $$\displaystyle 60^{\circ}$$ and $$\displaystyle 45^{\circ}$$, then the approximate distance between the objects is :

A
$$\displaystyle 50\left ( \sqrt{2}-2 \right )m$$

B
$$\displaystyle 50\left ( \sqrt{3}-3 \right )m$$

C
$$31 \ m$$

D
None of these

Solution

Let the two objects be C, D and the height of the tower is $$50$$m

In $$ΔABC$$

$$tan60^{0}=\dfrac{AB}{BC}$$

$$=\dfrac{50}{x}$$

$$\therefore x =\dfrac{ 50}{\sqrt 3}$$ ... {i}

Now, In $$ΔABD$$

$$tan45^{0}=\dfrac{AB}{BD}$$

$$\Rightarrow 1 =\dfrac{50}{x+y}$$

$$\Rightarrow x + y = 50$$

$$\Rightarrow y=x-50$$

$$\Rightarrow y=50 - \dfrac{50}{\sqrt 3}$$ ...[putting the value from {i}]

$$\Rightarrow y = \dfrac{50}{\sqrt 3}(\sqrt 3−1)$$

∴ Distance between two objects is $$=$$ $$\dfrac{50}{\sqrt 3}(\sqrt 3−1)$$

Rationalising the denominator we get the Distance$$=\dfrac{50 \sqrt{3}(\sqrt{3}-1)}{3}$$ m

Option D is correct.
Question 7
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The string of a kite $$100$$ metres long and it makes an angle of $$60^{\circ}$$ with the horizontal. Find the height of the kite.

A
$$86.5\text{ m}$$

B
$$90.5\text{ m}$$

C
$$80\text{ m}$$

D
$$75\text{ m}$$

Solution

Given, $$OK$$ is $$100\text{ m}$$ and the angle of elevation at O is $$60^{\circ}$$
Let the height of the kite be $$h$$
In right $$ΔOKP$$
$$\sin 60^{\circ}=\dfrac{h}{100}$$
$$\implies h=100 \sin 60^{\circ}$$
$$\implies h=100×\dfrac{\sqrt 3}{2}$$
$$\implies h=100×\dfrac { 1.732 }{ 2 } $$
$$\implies h=86.5 \text{ m}$$
Question 8
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The length of a ladder is exactly equal to the height of the wall it is leaning against. If the lower end of the ladder is kept on a bench of height 3 m. and the bench is kept 9 m. away from the wall, the upper end of the ladder coincides with the top of the wall. The height of the wall is

A
$$11 m$$

B
$$12 m$$

C
$$15 m$$

D
$$18 m$$

Solution

$$AB^2+BC^2=AC^2$$
$$(h-2)^2+9^2=h^2$$
$$h=15 m$$
Question 9
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The angle of depression of a car moving with uniform speed towards the building as observed from the top of the building is found to be $$30^{\circ}$$. The same angle of depression changes to $$60^{\circ}$$ after $$12$$ seconds. How much more time would the car take to reach the base?

A
$$6$$ sec

B
$$8$$ sec

C
$$4$$ sec

D
$$12$$ sec

Solution

Let $$'h'$$ be the height of the building.
BC $$=x$$ and CD $$=y$$
In $$\triangle ABC,$$
$$\Rightarrow \tan { 60° } = \dfrac { h }{ x } $$
$$\Rightarrow \sqrt { 3 } = \dfrac { h }{ x } $$
$$\Rightarrow h=\sqrt { 3 }x \longrightarrow \left( 1 \right) $$
In $$\triangle ADB, $$
$$ \Rightarrow \tan { 30° } =\dfrac { h }{ x+y } $$
$$ \Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ x+y } $$
$$ \Rightarrow x+y=\sqrt { 3 } h$$
From $$(1),$$
$$ \Rightarrow x+y=\sqrt { 3 } \times \sqrt { 3 } x$$
$$\Rightarrow x+y=3x$$
$$\Rightarrow y=2x$$
$$\Rightarrow x=\dfrac{y}{2}$$
Since $$y$$ is the distance traveled in $$12$$ sec from point D to point C.
so, the distance $$x$$ is travelled in $$\dfrac{12}{2}=6$$ sec. $$[\because$$ distance $$x$$ is half the distance $$y]$$
Hence, the answer is $$6$$ seconds
Question 10
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A tower is $$100\sqrt 3 m$$ high. The angle of elevation of its top from a point $$100 m$$ away from its foot is ........

A
$$60^o$$

B
$$45^o$$

C
$$30^o$$

D
$$22\frac {1}{2}^o$$

Solution

$$Ref.$$ Image
Let say BC is the tower and A is the point $$100m$$ away from its foot.
Let $$x$$ be the angle of elevation at A.
$$\therefore \tan x=\dfrac{BC}{AC}=\dfrac{100\sqrt{3}}{100}=\sqrt{3}$$

$$\therefore \tan x = \sqrt{3} = \tan60^o$$

$$\therefore x=60^{\circ}$$ (from trigonometric ratios)