# 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∘. Find the height of the tower.

26√3 m

24√3 m

#### SOLUTION

Solution :B

The above ΔABC represents the given situation.

AB = 24 m

In Triangle ABC,

tan60∘=√3=CBAB⇒AB×√3=CB⇒CB=24√3 m.

Hence, the height of the tower = 24√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∘ to the horizontal, would enable her to reach the required position, is

(You may take √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,

Sin 30∘=ABAC⇒12=8AC

Hence, height of the Ramp AC = 2 × 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∘. 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,

tan45∘=ABEBAlso, EB=DC∴tan45∘=ABDC⇒AB=DC×tan 45∘⇒AB=1×42.75

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∘ 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√3 m

12√3 m

#### SOLUTION

Solution :D

The given situation can be represented by the figure below

cos∠ACB=BCAC⇒cos30∘=BCAC⇒√32=12AC⇒AC=24√3=8×3√3=8√3mtan(∠ACB)=ABBC⇒tan30∘=AB12⇒AB=12√3=4√3m

∴ The height of tree was originally AB+AC = 4√3 +8 √3 = 12 √3m.

### 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∘. 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.

sin(∠ACB)=ABAC⇒sin30∘=ABAC⇒12×40=AB⇒AB=20m

### 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∘. Find the distance between the person and the foot of the tower.

30 m

20√3 m

20 m

60 m

#### SOLUTION

Solution :B

In ΔABC, tan(60∘) = BCAB

⇒ √3=60x

⇒ x = 20√3

⇒ Distance between the person and the tower = x = 20√3m.

### 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

tanθ=108=54

As both the shadows are formed at the same time, the sun rays forms the same angle in both the cases.

tanθ=h3054=h30

h = 37.5 cm

### Question 8

If the angle of elevation of a cloud from a point 200 m above a lake is 30∘ and the angle of depression of the reflection of the cloud in the lake from the same point is 60∘, then the height of the cloud above the lake is:

400 m

500 m

30 m

200 m

#### SOLUTION

Solution :A

In △ ABC',

tan60∘=x+400AB

√3=x+400AB

AB=x+400√3___(i)

In △ ABC,

tan30∘=xAB

1√3=xAB

AB=x√3 _______(ii)

Plugging the value of AB from equation (ii) in equation (i), we get

⇒x√3=x+400√3

⇒ 3x=x+400

⇒ 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∘ and 30∘. Find the distance between the two objects**. **(Take √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 ΔABC, tan 45∘ = ABAC = 1

AB=AC=100 m

In ΔABD, tan 30∘ = ABAD

AD×1√3=100

AD=100×√3=173.2m

CD=AD−AC=173.2−100=73.2 metres

### 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∘ , ∠ BCD = 45∘

tan30∘=BDBA⇒1√3=100BA⇒BA=100√3tan45∘=BDBC⇒1=100BC⇒BC=100

Distance between the two ships

=AC=BA+BC=100√3+100=100(√3+1)=100(1.73+1)=100×2.73=273m