# Free Applications of Trigonometry 02 Practice Test - 10th Grade

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.

A.

263 m

B.

243 m

C. 24 m
D. 26 m

#### SOLUTION

Solution : B The above ΔABC represents the given situation.

AB = 24 m

In Triangle ABC,
tan60=3=CBABAB×3=CBCB=243 m.

Hence, the height of the tower = 243 m.

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

___ m

(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

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?

A.

40 m

B.

35 m

C.

45 m

D.

50 m

#### SOLUTION

Solution : C

The given situation is represented by the figure below: In triangle ABE,
tan45=ABEBAlso, EB=DCtan45=ABDCAB=DC×tan 45AB=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.

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.

A.

12 m

B.

4 m

C.

43 m

D.

123 m

#### SOLUTION

Solution : D

The given situation can be represented by the figure below cosACB=BCACcos30=BCAC32=12ACAC=243=8×33=83mtan(ACB)=ABBCtan30=AB12AB=123=43m
The height of tree was originally AB+AC = 43 +8 3 = 12 3m.

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.

A.

10 m

B.

30 m

C.

20 m

D.

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)=ABACsin30=ABAC12×40=ABAB=20m

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.

A.

30 m

B.

203 m

C.

20 m

D.

60 m

#### SOLUTION

Solution : B In ΔABC, tan(60) = BCAB

3=60x

x = 203

Distance between the person and the tower = x = 203m.

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?

A.

65 cm

B.

75 cm

C.

37.5 cm

D.

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=h30

54=h30

h = 37.5 cm

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:

A.

400 m

B.

500 m

C.

30 m

D.

200 m

#### SOLUTION

Solution : A In ABC',
tan60=x+400AB
3=x+400AB
AB=x+4003___(i)

In ABC,
tan30=xAB
13=xAB
AB=x3 _______(ii)

Plugging the value of AB from equation (ii) in equation (i), we get
x3=x+4003
3x=x+400
x=200 m
Hence, the height of the cloud above the lake is
200 + 200 = 400 m

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)

A.

73.2 metres

B.

107.5 meters

C.

150 metres

D.

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×13=100
AD=100×3=173.2m
CD=ADAC=173.2100=73.2 metres

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.

A.

300 m

B.

173 m

C.

273 m

D.

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=BDBA13=100BABA=1003tan45=BDBC1=100BCBC=100

Distance between the two ships

=AC=BA+BC=1003+100=100(3+1)=100(1.73+1)=100×2.73=273m