Free Applications of Trigonometry 02 Practice Test - 10th Grade

Question 1

A tower stands vertically on the ground. From a point on the ground, which is 24 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^{\circ}$ . Find the height of the tower.

$26 \sqrt{3} m$

$24 \sqrt{3} m$

24 m

26 m

SOLUTION

Solution : B

The above $Δ A B C$ represents the given situation.

AB = 24 m

In Triangle ABC,
$t a n 60^{\circ} = \sqrt{3} = \frac{C B}{A B} \Rightarrow A B \times \sqrt{3} = C B \Rightarrow C B = 24 \sqrt{3} m .$

Hence, the height of the tower = $24 \sqrt{3} m .$

Question 2

A Technician has to repair a light on a pole of height 10 m. She needs to reach a point 2 m below the top of the pole to undertake the repair work. The length of the ramp that she should use which, when inclined at an angle of $30^{\circ}$ to the horizontal, would enable her to reach the required position, is

___ m

(You may take $\sqrt{3}$ = 1.73)

SOLUTION

Solution :

Let BD be the height of the pole. And A is the point where she has to do the repair work.

So, distance of the point she should reach = AB = 10 - 2 = 8 m
In the given right-angled triangle,
$S i n 30^{\circ} = \frac{A B}{A C}$

$\Rightarrow \frac{1}{2} = \frac{8}{A C}$
Hence, height of the Ramp AC = 2 $\times$ 8 = 16m

Question 3

An observer 2.25 m tall is 42.75 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is $45^{\circ}$ . What is the height of the chimney?

40 m

35 m

45 m

50 m

SOLUTION

Solution : C

The given situation is represented by the figure below:

In triangle ABE,
$tan 45^{\circ} = \frac{A B}{E B} A l s o, E B = D C ∴ t a n 45^{\circ} = \frac{A B}{D C} \Rightarrow A B = D C \times t a n 45^{\circ} \Rightarrow A B = 1 \times 42.75$

Hence, the height of the chimney = AC = AB + BC
We can observe that BC = ED.
Thus, AC = AB + ED
= 42.75 + 2.25
= 45 m.

Question 4

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^{\circ}$ with it. The distance between the foot of the tree to the point where the top touches the ground is 12 m. Find the height of the tree.

12 m

4 m

4 $\sqrt{3}$ m

12 $\sqrt{3}$ m

SOLUTION

Solution : D

The given situation can be represented by the figure below

$c o s ∠ A C B = \frac{B C}{A C} \Rightarrow c o s 30^{\circ} = \frac{B C}{A C} \Rightarrow \frac{\sqrt{3}}{2} = \frac{12}{A C} \Rightarrow A C = \frac{24}{\sqrt{3}} = 8 \times \frac{3}{\sqrt{3}} = 8 \sqrt{3} m t a n (∠ A C B) = \frac{A B}{B C} \Rightarrow t a n 30^{\circ} = \frac{A B}{12} \Rightarrow A B = \frac{12}{\sqrt{3}} = 4 \sqrt{3} m$
$∴$ The height of tree was originally AB+AC = $4 \sqrt{3} + 8 \sqrt{3} = 12 \sqrt{3} m .$

Question 5

A kite is flying on a string of 40 m. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $30^{\circ}$ . Find the height at which the kite is flying.

10 m

30 m

20 m

40 m

SOLUTION

Solution : C

The situation can be represented by the figure below:

AB is the height at which the kite is flying.
$s i n (∠ A C B) = \frac{A B}{A C} \Rightarrow s i n 30^{\circ} = \frac{A B}{A C} \Rightarrow \frac{1}{2} \times 40 = A B \Rightarrow A B = 20 m$

Question 6

A person observes the angle of elevation of the top of a 60m tower from a point on the level ground to be 60 $^{\circ}$ . Find the distance between the person and the foot of the tower.

30 m

20 $\sqrt{3}$ m

20 m

60 m

SOLUTION

Solution : B

In $Δ A B C$ , tan(60 $^{\circ}$ ) = $\frac{B C}{A B}$

$\Rightarrow$ $\sqrt{3} = \frac{60}{x}$

$\Rightarrow$ x = 20 $\sqrt{3}$

$\Rightarrow$ Distance between the person and the tower = x = 20 $\sqrt{3}$ m.

Question 7

A vertical stick 10 cm long casts a shadow 8 cm long. At the same time, a tower casts a shadow 30 cm long. Determine the height of the tower?

65 cm

75 cm

37.5 cm

100 cm

SOLUTION

Solution : C

$tan θ = \frac{10}{8} = \frac{5}{4}$

As both the shadows are formed at the same time, the sun rays forms the same angle in both the cases.
$tan θ = \frac{h}{30}$

$\frac{5}{4} = \frac{h}{30}$

h = 37.5 cm

Question 8

If the angle of elevation of a cloud from a point 200 m above a lake is $30^{\circ}$ and the angle of depression of the reflection of the cloud in the lake from the same point is $60^{\circ}$ , then the height of the cloud above the lake is:

400 m

500 m

30 m

200 m

SOLUTION

Solution : A

In $△$ ABC',
$t a n 60^{\circ} = \frac{x + 400}{A B}$
$\sqrt{3} = \frac{x + 400}{A B}$
$A B = \frac{x + 400}{\sqrt{3}}$ ___(i)

In $△$ ABC,
$t a n 30^{\circ} = \frac{x}{A B}$
$\frac{1}{\sqrt{3}} = \frac{x}{A B}$
$A B = x \sqrt{3}$ _______(ii)

Plugging the value of AB from equation (ii) in equation (i), we get
$\Rightarrow x \sqrt{3} = \frac{x + 400}{\sqrt{3}}$
$\Rightarrow 3 x = x + 400$
$\Rightarrow x = 200 m$
Hence, the height of the cloud above the lake is
200 + 200 = 400 m

Question 9

The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be $45^{\circ} a n d 30^{\circ}$ . Find the distance between the two objects. (Take $\sqrt{3} = 1.732$ )

73.2 metres

107.5 meters

150 metres

200 metres

SOLUTION

Solution : A

Let C and D be the objects and CD be the distance between the objects.

In $Δ A B C$ , $t a n 45^{\circ}$ = $\frac{A B}{A C}$ = 1

$A B = A C = 100 m$

In $Δ A B D$ , $t a n 30^{\circ}$ = $\frac{A B}{A D}$
$A D \times \frac{1}{\sqrt{3}} = 100$
$A D = 100 \times \sqrt{3} = 173.2 m$
$C D = A D - A C = 173.2 - 100 = 73.2 m e t r e s$

Question 10

Two ships are on either side of a 100 m lighthouse. The angles of elevation from the two ships to the top of the lighthouse are 30° and 45°. Find the distance between the ships.

300 m

173 m

273 m

200 m

SOLUTION

Solution : C

Let, BD be the lighthouse and A and C be the positions of the ships.

Then, BD = 100 m, $∠$ BAD = 30 $^{\circ}$ , $∠$ BCD = 45 $^{\circ}$

$t a n 30^{\circ} = \frac{B D}{B A} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{B A} \Rightarrow B A = 100 \sqrt{3} t a n 45^{\circ} = \frac{B D}{B C} \Rightarrow 1 = \frac{100}{B C} \Rightarrow B C = 100$

Distance between the two ships

$= A C = B A + B C = 100 \sqrt{3} + 100 = 100 (\sqrt{3} + 1) = 100 (1.73 + 1) = 100 \times 2.73 = 273 m$

Free Applications of Trigonometry 02 Practice Test - 10th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

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Question 8

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Question 9

SOLUTION

Question 10

SOLUTION

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