Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 47

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Question 1
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The angle of elevation of the top of an unfinished pillar at a point 150 m from its base is 30$$^\circ$$. If the angle of elevation at the same point is to be 45$$^\circ$$, then the pillar has to be raised to a height of how many metres?

A
59.4 m

B
61.4 m

C
62.4 m

D
63.4 m

Solution

In $$\Delta$$ABC
$$\displaystyle tan \, \theta \, = \, \frac{AB}{BC}$$
$$\displaystyle tan \, 30^{\circ} \, = \, \frac{AB}{150}$$
$$AB = 150$$ $$\displaystyle \times \, \frac{1}{\sqrt{3}} \,$$
$$ = \, 50 \sqrt{3} \, m$$
If $$\theta \, = \, 45^{\circ}$$
$$\displaystyle tan \, 45^{\circ} \, = \, \frac{AB}{BC}$$
$$ 1 \, = \, \dfrac{AB}{150}$$
$$AB = 150$$ m
Difference in height is $$ 150- 5\sqrt{3}$$
$$= 63.4$$m (approx.)
Question 2
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Two men standing on opposite sides of a flag staff measure the angles of the top of the flagstaff as $$30^{\circ}$$ and$$\, 60^{\circ}$$. If the height of the flagstaff is $$18$$ $$m$$ the distance between the men is :

A
$$24$$ $$m$$

B
$$\displaystyle24\sqrt{3}\,m$$

C
$$\displaystyle\frac{24}{\sqrt{3}}\,m$$

D
$$31.2$$ $$m$$

Solution

Let $$A$$ and $$B$$ be the points where the two men standing respectively.
Now let angle of elevation of the top of the flag-staff from $$A$$ be $$60^o$$ and that of, from the point $$B$$ be $$30^o$$.
Then from $$\Delta AOC$$
$$\tan 60^o=\dfrac{OC}{OA}$$
$$\Rightarrow \tan 60^o=\dfrac{18}{OA}$$
$$\Rightarrow OA=6\sqrt{3}$$.....(1).
And, from $$\Delta OBC$$
$$\tan 30^o=\dfrac{OC}{OB}$$
$$\Rightarrow \tan 30^o=\dfrac{18}{OB}$$
$$\Rightarrow OA=18\sqrt{3}$$.....(2).
Thus, the distance between the two men is $$AB=OA+OB$$$$=24\sqrt{3}$$ $$cm$$. [Using $$(1)$$ and $$(2)$$]
Question 3
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At a certain instant the ratio of the lengths of a pillar and its shadow are in the ratio $$1 : \sqrt{3}$$ at that instant, the angle of elevation of the sun is:

A
$$30^o$$

B
$$45^o$$

C
$$60^o$$

D
$$90^o$$

Solution

We can see that ABD is a Right Angled Triangle

it is given that,
$$\dfrac{AB}{BD}=\dfrac{1}{\sqrt3}\Rightarrow \tan\theta=\dfrac{1}{\sqrt3}\Rightarrow \theta=30^o$$

Therefore, Angle of Elevation of the Sun is $$30^o$$
Question 4
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Choose the correct answer from the alternatives given.
A tree is broken by the wind. If the top of the tree struck the ground at an angle of $$30^\circ$$ and at a distance of 30 m from the root, then the height of the tree is

A
$$15 \sqrt 3 m$$

B
$$20 \sqrt 3 m$$

C
$$25 \sqrt 3 m$$

D
$$30 \sqrt 3 m$$

Solution

In $$\Delta $$ ABC,
$$tan 30^{\circ} = \frac{BC}{AB}$$
$$\Rightarrow BC = 30 \times \frac{1}{\sqrt{3}} = 10\sqrt{3}$$m
and $$cos 30^{\circ} = \frac{AB}{AC}$$
$$\Rightarrow AC = \frac{AB}{\sqrt{3} / 2} = \frac{2 \times 30}{\sqrt{3}} = 20 \sqrt{3}$$
Height of the tree = BA = BC + CA
= 10$$\sqrt{3} + 10\sqrt{3} = 30\sqrt{3}$$m.
Question 5
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The angle of elevation of a tower from a point is $${ 30 }^{ \circ}$$. At a point on the horizontal line passing through the foot of the tower and $$50$$ metres nearer it, the angle of elevation is $${ 60 }^{ \circ }$$. The distance of the first point from the tower is

A
$$50\text{ m}$$

B
$$75\text{ m}$$

C
$$100\text{ m}$$

D
$$150\text{ m}$$

Solution

From $$\triangle ADB\Rightarrow \dfrac { AB }{ DB } =\tan{ 60 }^{ 0 }$$
$$\Rightarrow AB=50\sqrt { 3 } $$
also, $$\dfrac { AB }{ 50+x } =\tan{ 30 }^{ 0 }\Rightarrow \dfrac { 50+x }{ \sqrt { 3 } } =50\sqrt { 3 } $$
$$\Rightarrow \quad50+ x=150$$
$$\Rightarrow x=100 \text{ m}$$
Therefore $$CB=150\text{ m}$$
Question 6
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Two beams of length $${l}_{1}$$ and $${l}_{2}$$ are leaning on opposite sides of a thin vertical wall meeting at the same point on the wall and making angles $${30}^{o}$$ and $${60}^{o}$$ with it respectively. Then $${l}_{2}$$ is equal to

A
$$\cfrac { { l }_{ 1 } }{ 2 } $$

B
$$2{ l }_{ 1 }$$

C
$${ l }_{ 1 }\sqrt { 2 } $$

D
$${ l }_{ 1 }\sqrt { 3 } $$
Question 7
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A man on the top of a bamboo pole observes that the angle of depression of the base and the top of another pole are $${60}^{o}$$ and $${30}^{o}$$ respectively. If the second pole stands $$5$$ m above the ground level, then the height of the bamboo pole on which the man is sitting is

A
$$5$$ m

B
$$2.5$$ m

C
$$10$$ m

D
$$12.5$$ m

Solution

In $$\triangle ABC,\quad \cfrac { x }{ b } =\tan { { 30 }^{ o } } \Rightarrow b=x\sqrt { 3 } $$
In $$\triangle ADE\quad $$
$$\cfrac { x+5 }{ b } =\tan { { 60 }^{ o } } \Rightarrow \cfrac { x+5 }{ \sqrt { 3 } } =b$$
Equating we get, $$3x=x+5\Rightarrow x=\dfrac52$$
$$\therefore$$ height$$=2.5cm$$
Question 8
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The angles of elevation of the top of a temple, from the foot and the top of a building $$30\ m$$ high, are $$60^{\circ}$$ and $$30^{\circ}$$ respectively. Then height of the temple is

A
$$50\ m$$

B
$$43\ m$$

C
$$40\ m$$

D
$$45\ m$$

Solution

In $$\triangle DBC$$,

$$\tan 60^{\circ} = \dfrac {x + 30}{BC}$$

$$\Rightarrow BC = \dfrac {x + 30}{\sqrt {3}} ..... (1)$$

In $$\triangle DAE$$,

$$ \Rightarrow \tan 30^{\circ}=\dfrac{DE}{AE}$$

Now, $$AE = BC$$

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

$$\Rightarrow \sqrt {3}x = \dfrac {x + 30}{\sqrt {3}}$$

$$\Rightarrow 3x = x + 30$$

$$\Rightarrow 2x = 30$$

$$\Rightarrow x = 15\ m$$

$$\therefore\ h = x + 30 $$

$$= 15 + 30 = 45\ m$$
Question 9
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A 25 m long ladder is placed against a vertical wall such that the foot of the ladder is 7 m from the feet of the wall. If the top of the ladder slides down by 4 cm, by how much distance will the foot of the ladder slide?

A
$$9\ m$$

B
$$15\ m$$

C
$$5\ m$$

D
$$8\ m$$

E
$$4\ m$$
Question 10
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The angles of elevation of the top of a tower from two points A & B lying on the horizontal through the foot of the tower are respectively $$30^{\circ}$$ and $$45^{\circ}$$ of A & B are on the same side of the tower and AB = 48 m then the height of the tower is

A
24($$\sqrt{3} + 1)$$ m

B
25 m

C
26($$\sqrt{3} - 1)$$ m

D
23 m

Solution

Let the heightof tower be h.
$$tan 45^{\circ} = \cfrac{h}{BC}$$
$$l = \cfrac{h}{BC}$$
h = BC (1)
$$tan 30^{\circ} = \cfrac{h}{48 + BC}$$
$$\frac{1}{\sqrt{3}} = \cfrac{h}{48 + BC} \Rightarrow \cfrac{1}{\sqrt{3}} = \cfrac{h}{48 + h}$$ [From eq. (1)]
48+h= $$\sqrt{3}h \Rightarrow 48 = \sqrt{3}h -h$$
48 = h($$\sqrt{3} - 1) \Rightarrow h = \cfrac{48}{(\sqrt{3} - 1)} = 24(\sqrt{3} + 1)$$m.