Free Applications of Trigonometry 03 Practice Test - 10th Grade
Question 1
A kite is attached to a string of length 20√3m is tied to a pole on the level ground. If the string of the kite makes an angle of elevation of 60∘ with the ground, then the kite is flying at a height of 20 m.
True
False
SOLUTION
Solution : B
sin∠ACB=ABAC⇒sin60∘=ABAC⇒AB=AC × sin60∘=20√3×√32∴Height of the kite is 30m.
Question 2
A tower is 100√3 m high. Find the angle of elevation of its top from a point 100 m away from its foot.
SOLUTION
Solution : B
Let the angle of elevation be θ.
⇒ tan θ = PerpendicularBase
⇒ tan θ = 100√3100 = √3
Therefore, θ = 60∘
Question 3
The angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower is 30∘. Find the height of the tower.
10√3 m
20√3 m
30√3 m
20 m
SOLUTION
Solution : A
Angle of elevation = θ = 30∘
⇒ tanθ = PerpendicularBase
⇒ tanθ = h30
⇒ tan30∘ = 1√3
⇒ h30 = 1√3
Therefore, h=10√3m.
Question 4
The string of a kite is 100m long and it makes an angle of 60∘ with the horizontal. Find the height of the kite from the ground, assuming that there is no slack in the string.
100√3
200√3
50√3
100√3
SOLUTION
Solution : C
We know, angle of elevation = 60∘
⇒ sin θ = perpendicularhypotenuse
⇒ sin 60∘ = √32 = h100
⇒ h = 50√3 m.
Question 5
Find the angle of elevation of the sun when the shadow of a 10m long pole is 10√3 m.
30∘
60∘
45∘
15∘
SOLUTION
Solution : A
Let, angle of elevation of sun = θ
⇒ Tanθ = Perpendicularbase = 1010√3
⇒ Tanθ = 1√3
⇒ θ = 30∘
Question 6
A circus artist is climbing a 20m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. The height of the pole is
m
SOLUTION
Solution :
Angle of elevation θ = 30∘
⇒ Sinθ = perpendicularhypotenuse
⇒ Sin30∘ = 12 = height of pole20
Thus, height of pole = 10 m.
Question 7
Two towers A and B are standing at some distance apart. From the top of tower A, the angle of depression of the foot of tower B is found to be 30∘. From the top of tower B, the angle of depression of the foot of tower A is found to be 60∘. If the height of tower B is ‘h’ m then the height of tower A in terms of ‘h’ is _____ m
h2m
h3m
√3hm
h√3m
SOLUTION
Solution : B
Let the height of tower A be = AB = H.
And the height of tower B = CD = h
In triangle ABC
tan30∘ = ABAC = HAC ...... (1)
In triangle ADC
tan60∘ = CDAC = hAC...... (2)
Dividing (1) by (2) we get, tan30∘tan60∘ = Hh
H= h3
Question 8
The angle of elevation of the top of a tree from a point A on the ground is 60∘ . On walking 20 m away from point A, to a point B, the angle of elevation changes to 30∘. Find the height of the tree.
10√3 m
20√3 m
30√3 m
40√3 m
SOLUTION
Solution : A
In triangle ACD,
tanθ=DCCA=hx
⇒tan60∘=√3=hx ..................... 1
In triangle CDB,
⇒tan30∘=CDCB
⇒1√3=hx+20....................... 2
Using (1) and (2), we get⇒1√3=x√3x+20
⇒x+20=3x
⇒2x=20
⇒x=10
∴h=x√3=10√3
Thus, the height of the tree is 10√3 m.
Question 9
The tops of two poles of height 14 m and 20 m are connected by a wire which makes an angle of 30° with the horizontal. The length of the wire is _____m.
6m
10m
8m
12m
SOLUTION
Solution : D
In triangle ABC,
BC = y = 20-14 = 6m;
Let AB = x
⇒ Sin30∘= BCAB
⇒ 12 = 6x
⇒ x = 12
Thus, the length of the string is 12 m.
Question 10
From the top of a cliff 25m high the angle of elevation of a tower is found to be equal to the angle of depression to the foot of the tower. The height of the tower is ___.
25m
75m
50m
100m
SOLUTION
Solution : C
In triangle ABE,
tanθ2 = ABBE = 25BE
In triangle ADC,
tanθ1 = CDAD
We know, θ1 = θ2
⇒ tanθ1 = tanθ2
CDAD = 25BE
⇒ CD = 25 [Since AD = BE]
DE =AB = 25m⇒ Height of tower = CD+ DE
= 25+25 = 50m