Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 28

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Question 1
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A boy $$1.7$$ m tall, is $$25$$ m away from a tower and observes the angle of elevation of the top of the tower to be $${60}^{o}$$. Find the height of the tower.

A
$$55$$ m

B
$$45$$ m

C
$$20$$ m

D
$$35$$ m

Solution

Let $$AA'$$ be the boy and the tower be $$B'C$$
Therefore, $$AA'=1.7$$ m $$=BB'$$
and $$A'B'=25\ m=AB$$
Now, angle of elevation is $$\angle CAB=\theta$$.
Given, $$\theta=60^o $$
$$\Rightarrow \tan\theta=\dfrac{BC}{AB}$$
$$\Rightarrow 25\tan 60^o=BC$$
$$\Rightarrow BC=25\sqrt3$$
Now, height of tower
$$=BC+BB'$$
$$=25\sqrt3+1.7$$
$$=45$$ m
Question 2
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A tree is broken by the wind. Its top struck the ground at an angle $${30}^{o}$$ at a distance of $$30m$$ from its foot. The whole height of the tree is:

A
$$10\sqrt 3\ m$$

B
$$20\sqrt 3\ m$$

C
$$40\sqrt 3\ m$$

D
$$30\sqrt 3\ m$$

Solution

From figure,

In $$\triangle ABC$$

$$\tan 30 = \dfrac{BC}{AB}=\dfrac{1}{\sqrt 3}$$

$$\therefore BC=\dfrac{30}{\sqrt 3}$$

$$\cos 30 = \dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$$

$$\therefore AC=\dfrac{60}{\sqrt 3}$$

Total height,

$$=BC+AC$$

$$=\dfrac{30}{\sqrt 3}+\dfrac{60}{\sqrt 3}$$

$$=30\sqrt{3}\ m$$
Question 3
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The height of a tower is h and the angle of elevation of the top of the tower is $$\displaystyle \alpha $$ on moving a distance h/2 towards the tower, the angle of elevation becomes. $$\displaystyle \beta $$. What is the value of$$\quad \cot { \alpha } -\cot { \beta } \quad $$

A
$$\displaystyle \dfrac{1}{2}$$

B
$$\displaystyle \dfrac{2}{3}$$

C
$$1$$

D
$$2$$

Solution

Let $$AO=h$$ be the tower O being the foot of it.

Then $$\quad \angle OCA=\alpha \quad \& \quad \angle OBA=\beta ,\\ AO=h,\\ OC-OB=BC=\dfrac { h }{ 2 } \\ \therefore \quad \dfrac { OC }{ OA } =\cot { \alpha } \quad and\quad \dfrac { OB }{ OA } =\cot { \beta } .\\ \therefore \quad \cot { \alpha } -\cot { \beta } \\ =\dfrac { OC }{ OA } -\dfrac { OB }{ OA } \\ =\dfrac { OC-OB }{ AO } \\ =\dfrac { BC }{ AO } =\dfrac { \dfrac { h }{ 2 } }{ h } =\dfrac { 1 }{ 2 } .\quad $$
Question 4
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The ladder resting against a vertical wall is inclined at an angle of $${30}^{\circ}$$ to the ground. The foot of the ladder is $$7.5\text{ m}$$ from the wall. Find the length of the ladder.

A
$$8.66\text{ m}$$

B
$$7.66\text{ m}$$

C
$$10.66\text{ m}$$

D
$$9.66\text{ m}$$

Solution

Let the length of the ladder be $$\lambda .$$

In $$\triangle ABC,$$
$$\begin{aligned}{}\cos 30^\circ& = \frac{{7.5}}{\lambda}\\\frac{{\sqrt 3 }}{2} &= \frac{{7.5}}{\lambda}\\\lambda&= \frac{{15}}{{\sqrt 3 }}\\ &= 5\sqrt 3\\&\approx 8.66 \text{ m}\end{aligned}$$

Hence, the length of the ladder is $$8.66\text{ m}.$$
Question 5
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What is the angle of elevation of a vertical flagstaff of height $$100\sqrt 3m$$ from a point $$100m$$ from its foot.

A
$${18}^{o}$$

B
$${30}^{o}$$

C
$${48}^{o}$$

D
$${60}^{o}$$

Solution

Height of flagstaff (H) = $$100 \sqrt{3}$$
Distance between the point and the flagstaff (D) = 100 m
Let the angle of elevation be $$\theta$$
Then,

$$\tan \theta = \dfrac{D}{H}$$
$$\tan \theta = \dfrac{100\sqrt{3}}{100}$$
$$\tan \theta = \sqrt{3}$$
$$\tan \theta = \tan 60^{\circ}$$
$$\theta = 60^{\circ}$$
Question 6
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A ladder $$20m$$ long is placed against a vertical wall of height $$10$$ metres. Find the distance between the foot of the ladder and the wall and also the inclination of the ladder to the horizontal.

A
$$17.32m$$; $$\theta={30}^{o}$$

B
$$17.32m$$; $$\theta={60}^{o}$$

C
$$17.32m$$; $$\theta={45}^{o}$$

D
$$27.32m$$; $$\theta={30}^{o}$$

Solution

Since, Hypotenuse$$\,=20m$$
and Height=$$\,10m$$
$$\therefore sin\theta\,$$$$=\dfrac{10}{20}$$

$$\Rightarrow sin\theta\,$$$$=\dfrac{1}{2}$$
But we know that
$$sin30^{o}=\,$$$$\dfrac{1}{2}$$

Hence, we can say that
$$\theta=30^{o}$$
Therefore,
Base
$$=20cos30^{o}m$$
$$=20(\frac{\sqrt{3}}{2})$$
$$=10(1.732)$$
$$=17.32m$$
Question 7
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The angle of elevation of the top of a tower from a distance $$100m$$ from its foot is $${30}^{o}$$. The height of the tower is:

A
$$100\sqrt 3m$$

B
$$\cfrac {200}{\sqrt 3}m$$

C
$$50\sqrt 3 m$$

D
$$\cfrac{100}{\sqrt 3}m$$

Solution

Given Angle Of elevation $$30^{\circ}$$

The distance from its foot is $$100m$$

Consider $$\tan $$ of the given angle

$$\implies \tan \theta =\dfrac {Height}{Distance}$$

$$\implies \tan 30=\dfrac {Height}{100}$$

$$\implies \dfrac 1{ \sqrt 3} =\dfrac {Height}{100}$$

$$\implies Height=\dfrac {100}{\sqrt 3}$$
Question 8
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A tower is $$120\ m$$ high. Its shadow is $$x\ m$$ shorter when the sun's altitude is $$\displaystyle 60^{\circ}$$ than when it was $$45^{\circ}$$. Find $$x$$ correct to nearest metre

A
$$51\ m$$

B
$$42\ m$$

C
$$26\ m$$

D
None of the above

Solution

Let $$AB$$ be the tower and its shadows be $$BD$$ and $$BC$$ corresponding to the angle of elevations $$60^{\circ}$$ and $$45^{\circ}$$ respectively

In $$\Delta ABC,$$
$$ \tan 45^{\circ}=\dfrac{AB}{BC}$$

$$\Rightarrow 1=\dfrac{120}{BC}$$

$$\Rightarrow BC=120\:m$$

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

$$\Rightarrow \sqrt{3}=\dfrac{120}{BD}$$

$$\Rightarrow BD=40\sqrt{3}\ m$$

Now,
$$\begin{aligned}{}BC &= BD + CD\\120 &= 40\sqrt 3 + x\\x &= 120 - 40\sqrt 3 \\ &= 40\sqrt 3 (\sqrt 3 - 1)\\& \approx 51\;m\end{aligned}$$
Question 9
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A 10 m long flagstaff is fixed on the top of a tower from a point on the ground the angles of elevation of the top and bottom of flagstaff are $$\displaystyle 45^{\circ}$$ and $$\displaystyle 30^{\circ}$$ respectively. Find the height of the tower.

A
$$13.66\ m$$

B
$$18.22\ m$$

C
$$22\ m$$

D
$$16\ m$$

Solution

Let $$AB$$ be the flagstaff and $$BC$$ be the tower

Let $$CD = x$$ m and $$BC = h$$ m

In $$\displaystyle \Delta ACD,$$

$$\tan 45^{\circ}=\dfrac{AC}{DC}$$

$$=\dfrac{h+10}{x}$$

$$\displaystyle \Rightarrow 1=\frac{h+10}{x}$$

$$\Rightarrow x=h+10$$ $$...(i)$$

In $$\displaystyle \Delta BCD,$$

$$\tan 30^{\circ}=\dfrac{BC}{DC}$$

$$=\dfrac{h}{x}$$

$$\displaystyle \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{h}{x}$$

$$\Rightarrow x=h\sqrt{3}=1.732h$$ $$...(ii)$$

From $$(i)$$ and $$(ii),$$

$$\Rightarrow 1.732 h = h + 10$$

$$\Rightarrow 0.732h=10$$

$$\Rightarrow h=\dfrac{10}{0.732}=13.66\: m$$
Question 10
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The angle of depression of $$47$$ m high building from the top of a tower $$137$$ m high is $$\displaystyle 30^{\circ}$$. Calculate the distance between the building and the tower

A
$$\displaystyle 90\sqrt{3}$$ m

B
$$\displaystyle 80\sqrt{5}$$ m

C
$$\displaystyle 70\sqrt{3}$$ m

D
None of these

Solution

Let $$AB$$ and $$CD$$ represent the tower and building respectively
The angle of depression $$\displaystyle \angle XAD=30^{\circ}$$
In $$\displaystyle \Delta ADE,DE=CB=x$$ $$\displaystyle \angle ADE=\angle XAD=30^{\circ}(alt.\angle s)$$
$$AE = AB-EB = AB - DC = (137 - 47) m = 90$$ m
$$\displaystyle \therefore \tan 30^{\circ}=\frac{AE}{DE}$$
$$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{90}{x}$$
$$\Rightarrow x=90\sqrt{3}$$ m