Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 54

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Question 1
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The angle of elevation of the top of a tower at a distance of 30 m from its foot is $$60^\circ$$. The height of the tower is

A
$$20$$m

B
$$30\sqrt{3}$$m

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

D
$$\cfrac{15}{\sqrt{2}}$$m

Solution

Let the height of the tower be $$AB$$ and the point be $$C$$

Hence $$BC=30\ m$$

and $$\angle ACB=60^\circ$$

In $$\triangle ABC,$$

$$\tan 60^\circ=\dfrac{AB}{AC}$$

$$\Rightarrow {\sqrt 3}=\dfrac{AB}{30}$$

$$\therefore AB=30\sqrt 3\ m$$
Question 2
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The shadow of a tower of height $$(1+\sqrt { 3 } )$$ m standing on the ground is found to be 2 m longer when the sun's elevation is $$30^0$$, then when the sun's elevation was

A
$$30^0$$

B
$$45^0$$

C
$$60^0$$

D
$$75^0$$

Solution

From given, we have,

$$\tan 30 =\dfrac{1+\sqrt 3}{x+2}$$

$$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{1+\sqrt 3}{x+2}$$

$$\Rightarrow x+2=\sqrt 3+3$$

$$\therefore x=1+\sqrt 3$$

now,

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

$$\Rightarrow \tan \theta =1$$

$$\therefore \theta =\tan ^{-1}1=45^{\circ}$$
Question 3
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The angle of elevations of the top of a tower from the top and bottom of a building of height a are $${ 30 }^{ \circ }$$ and $${ 45 }^{ \circ }$$ respectively, If the tower and the building stand at the same level, then the height of the tower is :

A
$$a\sqrt { 3 } $$

B
$$\cfrac { a\sqrt { 3 } }{ \sqrt { 3 } -1 } $$

C
$$\cfrac { a(3+\sqrt { 3 } ) }{ 2 } $$

D
None of these

Solution
Question 4
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The top of a broken tree touches the ground at a distance of $$12 \,m$$ from its base. If the tree is broken at a height of $$5 \,m$$ from the ground hen the actual height of the tree is

A
$$25 \,m$$

B
$$13 \,m$$

C
$$18 \,m$$

D
$$17 \,m$$

Solution

Let $$AB$$ is the unbroken part of the tree and $$BC$$ is the broken part of it.
Here, triangle ABC is right-angled triangle.
$$\Rightarrow(BC)^{2}=(AB)^{2}+(AC)^{2}$$
$$\Rightarrow(BC)^{2}=5^{2}+12^{2}$$
$$\Rightarrow(BC)^{2}=25+144$$
$$\Rightarrow(BC)^{2}=169$$
$$\Rightarrow(BC)^{2}=13m$$
Now actual height of tree $$=AB+BC$$
$$\Rightarrow 5+13$$
$$\Rightarrow 18m$$
Question 5
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The tops of two poles of height $$18$$m and $$10$$m are connected by wire. If the wire makes an angle of measure $$45^o$$ with the horizontal, then the length of wire is _________.

A
$$16$$m

B
$$18$$m

C
$$8\sqrt{2}$$m

D
$$8$$m

Solution

Let $$AB$$ be the pole having height $$18\ m$$ and $$CD$$ be the pole of height $$10\ m$$

$$AE=AB-EB$$ $$[DC=EB]$$

$$AE=18-10=8\ m$$

In $$\triangle ADE,$$

$$\sin 30^\circ=\dfrac{AE}{AD}$$

$$\Rightarrow \dfrac{1}{2}=\dfrac{8}{AD}$$

$$\Rightarrow AD=2\times 8$$

$$\Rightarrow AD=16\ m$$

Hence, the length of the wire is $$16\ m$$
Question 6
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The ratio of the length of a tree and its shadow is $$ 1:
\frac{1}{\sqrt{3}} $$. The angle of the sun's elevation is

A
$$ 30^o $$

B
$$ 45^o $$

C
$$ 60^o $$

D
$$ 90^o $$

Solution
Question 7
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Find the angle of elevation of the top of a tower, whose height is $$ 100 m$$, at a point whose distance from the base of the tower is $$ 100 m$$.

A
$$ 30^o $$

B
$$ 60^o $$

C
$$ 45^o $$

D
$$ 90^o $$

Solution
Question 8
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Choose the correct alternative answer for the following question.
When we see at a higher level, from the horizontal line, angle formed is .............

A
angle of elevation

B
angle of depression

C
0

D
straight angle

Solution

When we see at a higher level, from the horizontal line, angle formed is angle of elevation.

Hence, the correct answer is angle of elevation.
Question 9
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If the shadow of 10 m high tree is $$ 10 \sqrt{3} \ m $$, then find the angle of elevation of the sun.

A
$$ 60^o $$

B
$$ 90^o $$

C
$$ 45^o $$

D
$$ 30^o $$

Solution

We can draw the figure shown above, based on the given infomation.
From $$\triangle ABC$$,
$$AB =$$ height of tree $$= 10\ m$$
$$BC =$$ shadow of tree

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

Hence, $$\tan \theta = \dfrac {AB}{BC}$$

$$\Rightarrow \tan \theta = \dfrac {10}{10\sqrt {3}} $$

$$= \dfrac {1}{\sqrt {3}}$$

$$\Rightarrow \tan \theta = \tan 30^{\circ}\quad \quad \left[\because \ \tan 30^o=\dfrac{1}{\sqrt3}\right]$$

$$\Rightarrow \theta = 30^{\circ}$$

Hence, the angle of the elevation of the Sun is $$30^\circ$$.
Question 10
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The shadow length of a vertical pillar is same as the height of pillar, then angle of elevation of sun is___.

A
$$45^{\circ}$$

B
$$30^{\circ}$$

C
$$60^{\circ}$$

D
$$50^{\circ}$$

Solution

Let $$BC$$ be a tower with height $$h\ m$$.
Then according to question, its shadow length $$AB$$ will be $$h\ m$$.

Let angle of elevation $$= \theta$$

In right angled $$\Delta ABC$$, we have
$$\tan \theta =\dfrac{BC}{AB} $$
$$\Rightarrow \tan\theta=\dfrac{h}{h} $$
$$\Rightarrow \tan\theta=1$$
$$\Rightarrow \tan \theta = \tan 45^{\circ}$$
$$\Rightarrow \theta = 45^{\circ}$$

Hence, the angle of elevation is $$45^{\circ}$$.