Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 23

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Question 1
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If a pole of height 6 m casts a shadow $$2\sqrt{3}$$ long on the ground, then the suns elevation is.

A
$$30^{\circ}$$

B
$$60^{\circ}$$

C
$$45^{\circ}$$

D
$$90^{\circ}$$

Solution

as we can observe that,
The Angle of Elevation of Sun at point $$P$$ is equal to the Angle of Elevation of the Building at point $$P.$$

so, $$\tan A=\dfrac{6}{b}=\dfrac{6}{2\sqrt3}=\sqrt3\Rightarrow A=60^o$$

Therefore, Answer is $$60^o$$
Question 2
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The angle of elevation of the top of the tower from a point on the ground, which is $$30 m$$ away from the foot of the tower is $$45^{\circ}$$. The height of the tower (in metres) is:

A
$$15$$

B
$$30$$

C
$$30 \sqrt{3}$$

D
$$10 \sqrt{3}$$

Solution

Let the height of tower be $$'h'$$ $$m.$$
Now, in $$\triangle ABC,$$
$$\Rightarrow \tan { 45^{\circ} } =\dfrac { AB }{ BC } =\dfrac { h }{ 30 } $$
$$\Rightarrow 1=\dfrac { h }{ 30 } $$
$$\Rightarrow h=30m$$
Hence, the answer is $$30m.$$
Question 3
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A Lamp Post, $$5\sqrt{3}m$$ high, cast a shadow $$5\ m$$ long on the ground. The suns elevation of this point is:

A
$$30^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
$$90^{\circ}$$

Solution

Let the angle of elevation be $$'\theta'.$$
In, $$\triangle ACB,$$
$$\Rightarrow \tan { \theta } =\dfrac { AB }{ BC } =\dfrac { 5\sqrt { 3 } }{ 5 } $$
$$\Rightarrow \tan { \theta } =\sqrt { 3 } $$
$$\Rightarrow$$ Or $$ \theta =60^{\circ}$$
Hence, the answer is $$60^{\circ}.$$
Question 4
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When the angle of elevation of the sun is $$ 30^{\circ}$$, the length of the shadow cast by 50 m high building is

A
$$\dfrac{50}{\sqrt{3}}m$$

B
$$50\sqrt{3}m$$

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

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

Solution

Let the length of shadow be $$'x'\ m$$
In $$ACB,$$
$$\Rightarrow \tan { 30^{\circ} } =\dfrac { AB }{ BC } =\dfrac { 50 }{ x } $$
$$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { 50 }{ x } $$
$$\Rightarrow x=50\sqrt { 3 } m$$
Hence, the answer is $$50\sqrt { 3 } m.$$
Question 5
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A tower stands vertically on the ground. From a point on the ground which is $$25\ m$$ away from the foot of the tower, the angle of elevation of the top of the tower is found to be $$45^{\circ}$$. Then the height $$(in\ meters)$$ of the tower is:

A
$$25 \sqrt{2}\ m$$

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

C
$$25\ m$$

D
$$12.5\ m$$

Solution

Let the height of the tower be $$'h'$$ $$m.$$
Now, in $$\triangle ACB,$$
$$\Rightarrow \tan { 45^{\circ}} =\dfrac { AB }{ BC } =\dfrac { h }{ 25 } $$
$$\Rightarrow 1=\dfrac { h }{ 25 } $$
$$\Rightarrow h=25\ m$$
Hence, the height of tower is $$25\ m.$$
Question 6
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Directions For Questions

From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be $$\alpha $$ and $$\beta $$, where $$\alpha >\beta $$.

...view full instructions

The height of the tower if distance between the objects, $$p=150\ m,\alpha =60^o\: and\: \beta =30^o$$, is:

A
$$120\ m$$

B
$$130\ m$$

C
$$140\ m$$

D
none of the above

Solution

$$Height \: of \: the \: tower \: (AB)=h\:m$$
$$distance \: (CD) =p \: m$$
$$Let \:distance (BC) =xm$$
$$\angle ACB=\alpha \: and\: \angle ADB=\beta $$
In right $$\triangle ABD$$,
$$\frac{AB}{BC}=\tan \alpha $$
$$\Rightarrow \frac{h}{x}=\tan \alpha $$
$$\Rightarrow {h}={x}\tan \alpha $$ .....( i)
In right $$\triangle ABD$$,
$$\frac{AB}{BD}=\tan \beta $$
$$\Rightarrow \frac{h}{BC+CD}=\tan \beta$$
$$\Rightarrow h=(x+p)\tan \beta $$ ..... (ii)
From (1), we get
$$x=\frac{h}{\tan \alpha }$$
Thus, $$h = (\frac{h}{\tan\alpha} + p) \tan \beta$$
$$h = (\frac{h + p \tan \alpha}{\tan \alpha}) \tan \beta$$
$$h \tan \alpha = h \tan \beta + p \tan \alpha \tan \beta$$
$$h (\tan \alpha - \tan \beta) = p \tan \alpha \tan \beta$$
$$h = \frac{p \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}$$

Now, putting $$p=150\:m,\alpha =60^o\: and\: \beta =30^o$$, we get
$$h=\frac{150\times \tan 30^o\times \tan 60^o }{\tan 60^o-\tan 30^o }m=129.9\: m$$
Hence, $$the \: height \: of \: the \: tower =129.9\:m\simeq 130\:m$$
Question 7
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Directions For Questions

From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be $$\alpha $$ and $$\beta $$ where $$\alpha >\beta $$.

...view full instructions

If the distance between the objects is $$p$$ metres, then the height $$h$$ of the tower is:

A
$$\dfrac{p\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$$

B
$$\dfrac{\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$$

C
$$\dfrac{p(\tan \alpha -\tan \beta )}{\tan \alpha \tan \beta }$$

D
none of the above

Solution

$$Height \: of \: the \: tower \: (AB)=h\:m$$
$$distance \: (CD) =p \: m$$
$$Let \:distance (BC) =xm$$
$$\angle ACB=\alpha \: and\: \angle ADB=\beta $$
In right $$\triangle ABD$$,
$$\frac{AB}{BC}=\tan \alpha $$
$$\Rightarrow \frac{h}{x}=\tan \alpha $$
$$\Rightarrow {h}={x}\tan \alpha $$ .....( i)
In right $$\triangle ABD$$,
$$\frac{AB}{BD}=\tan \beta $$
$$\Rightarrow \frac{h}{BC+CD}=\tan \beta$$
$$\Rightarrow h=(x+p)\tan \beta $$ ..... (ii)
From (1), we get
$$x=\frac{h}{\tan \alpha }$$
Thus, $$h = (\frac{h}{\tan\alpha} + p) \tan \beta$$
$$h = (\frac{h + p \tan \alpha}{\tan \alpha}) \tan \beta$$
$$h \tan \alpha = h \tan \beta + p \tan \alpha \tan \beta$$
$$h (\tan \alpha - \tan \beta) = p \tan \alpha \tan \beta$$
$$h = \frac{p \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}$$
Hence proved.
Question 8
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If the length of the shadow of a tower is $$\sqrt{3}$$ times that of its height, then the angle of elevation of the sun is

A
$$15^o$$

B
$$30^o$$

C
$$45^o$$

D
$$60^o$$

Solution

Let height of tower (AB) be $$h\ \text{metres}$$
Then length of its shadow will be $$BC = \sqrt{3}\:h\:\text{metres}$$
Let angle of elevation be $$\theta $$
Now, In $$\triangle ABC$$
$$\tan \theta = \dfrac{P}{B} = \dfrac{AB}{BC}$$
$$\tan \theta = \dfrac{h}{\sqrt{3}h}=\dfrac{1}{\sqrt{3}}$$
$$\Rightarrow \theta =30^o$$
Question 9
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The top of a broken tree has it's top touching the ground (shown in the adjoining figure) at a distance of $$10\ m$$ from the bottom. If the angle made by the broken part with the ground is $$30^o$$, then the length of the broken part is:

A
$$10\sqrt{3}\: m$$

B
$$\dfrac{20}{\sqrt{3}}\:m$$

C
$$20\ m$$

D
$$20\sqrt{3}\:m$$

Solution

The tree broke at point B, and touches the ground at A. The roots are at point C.
Now, the distance between the top of broken part and the roots $$= 10\ m$$
Considering it as a triangle system,$$\triangle ABC$$ is a right angled triangle.
$$\displaystyle \cos 30^o = \frac{Perpendicular}{Hypotenuse} = \frac{AC}{AB}$$
$$\displaystyle \frac{\sqrt{3}}{2} = \frac{10}{AB}$$
$$\therefore$$ Length of broken part $$\displaystyle AB = \frac{20}{\sqrt{3}}\ m$$
Question 10
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A tree breaks due to storm and the broken part bends so that the top of the tree just touches the ground, making an angle of $$30^o$$ with the horizontal. The distance from the foot of the tree to the point where the top touches the ground is 10 m. The height of the tree is

A
$$10(\sqrt 3+1)\ m$$

B
$$10\sqrt 3\ m$$

C
$$10(\sqrt 3-1)\ m$$

D
$$\frac {10}{\sqrt 3}\ m$$

Solution

Let BD be the tree broken at point C such that the broken part CD takes the position CA and strikes the ground at A. It is given that $$AB=10\ \text{m}$$ and $$\angle{BAC=30^{\circ}}$$
Let $$BC=x\ \text{m}$$ and $$CD=CA=y\ \text{m}$$
In right angled triangle ABC, we have
$$\tan30^{\circ}=\dfrac{BC}{AB}$$
$$\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{x}{10}$$
$$\Rightarrow x=\dfrac{10}{\sqrt3}=\dfrac{10\sqrt3}{3}$$
And $$\cos30^{\circ}=\dfrac{AB}{AC}$$
$$\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{10}{y}$$
$$\Rightarrow y=\dfrac{20}{\sqrt3}=\dfrac{20\sqrt3}{3}$$
Now,
Height of the tree $$=x+y$$
$$=\dfrac{10\sqrt3}{3}+\dfrac{20\sqrt3}{3}$$
$$=\dfrac{30\sqrt3}{3}$$
$$=10\sqrt3\ \text{m}$$

Hence, the height of the tree is $$10\sqrt3\ \text{m}$$.