Trigonometry Ex 6.2-10th Std Maths-Book Back Question And Answer
Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 3–√ m.
Answer:
Height of the tower (AC) = 10 3–√ m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ
In the right ∆ ABC, tan θ = ACAB
= 103√30=3√3
tan θ = 13√ = tan 30°
∴ Angle of inclination is 30°
Question 2.
A road is flanked on either side by continuous rows of houses of height 43–√ m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the width of the road.
Answer:
Let the mid point of the road AB is “P” (PA = PB)
Height of the home = 43–√ m
Let the distance between the pedestrian and the house be “x”
In the right ∆ APD, tan 30° = ADAP
13√=43√x
x = 4 3–√ × 3–√ = 12 m
∴ Width of the road = PA + PB
= 12 + 12
= 24 m
Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (3–√ = 1.732)
Answer:
Let the height of the window FE be “h” m
Let FC be “x” m
∴ EC = (h + x) m
In the right ∆ CDF, tan 45° = CECD
1 = x5 ⇒ x = 5
In the right ∆ CDE, tan 60° = CECD
3–√ = x+h5 ⇒ x + h = 53–√
5 + h = 5 3–√ (substitute the value of x)
h = 5 3–√ – 5 = 5 × 1.732 – 5 = 8. 66 – 5 = 3.66
∴ Height of the window = 3.66 m
Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40° . Find the height of the pedestal.
(tan 40° = 0.8391, 3–√ = 1.732)
Answer:
Height of the statue = 1.6 m
Let the height of the pedestal be “h”
AD = H + 1.6m
Let AB be x
In the right ∆ ABD, tan 60° = ADAB
3–√ = h+1.6x
x = h+1.63√ ……..(1)
In the right ∆ ABC, tan 40° = ACAB
0.8391 = hx
x = h0.8391
Substitute the value of x in (1)
h0.8391=h+1.63√
(h + 1.6) 0.8391 = 3–√ h
0.8391 h + 1.34 = 1.732 h
1.34 = 1.732 h – 0.8391 h
1.34 = 0.89 h
h = 1.340.89 = 13489 = 1.5 m
Height of the pedestal = 1.5 m
Question 5.
A Flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (3–√ = 1.732)
Answer:
Height of the Flag pole (ED) = h m
AF and AD is the radius of the semi circle (r)
AC = (r + 7)
AB = (r + 7 + 5)
= (r + 12)
In the right ∆ ABD, tan 30° = ADAB
13√=rr+12
3–√ r = r + 12
3–√ r – r = 12 ⇒ r (3–√ – 1) = 12
r[1.732 – 1] = 12 ⇒ 0.732 r = 12
r = 120.732 ⇒ = 16.39 m
In the right ∆ ACE, tan 45° = AEAC
1 + r+hr+7
r + 7 = r + h
∴ h = 7 m
Height of the pole (h) = 7 m
Radius of the dome (r) = 16.39 m
Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer:
Let the height of the electric pole AD be “h” m
EC = 15 – h m
Let AB be “x”
In the right ∆ ABC, tan 60° = BCAB
3–√ = 15x
x = =153√=15×3√3
= 53–√
In the right ∆ CDE, tan 30° = ECDE
13√=15−hx ………….(1)
Substitute the value of x = 5 3–√ in (1)
13√=15−h53√⇒3–√(15−h)=53–√
(15 – h) = 53√3√ ⇒ 15 – h = 5
h = 15 – 5 = 10
∴ Height of the electric pole = 10 m
Question 7.
A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Answer:
Let the first part of the pole be “x” and the second part be “9x”
∴ height of the pole (AC) = x + 9x = 10x
Given ∠CDB = ∠BDA
∴ BD is the angle bisector of ∠ADC
By angle bisector theorem
ABBC = ADDC
9xx = AD25 ⇒ AD = 9 × 25 = 225
In the right ∆ ACD
AD2 = AC2 + CD2
(225)2 = (10x)2 + 252
50625 = 100x2 + 625
∴ 100x2 = 50625 – 625 = 50000
x2 = 50000100 = 500
x = 500−−−√ = 5×100−−−−−−√=105–√
∴ AC = 10 × 105–√ = 100 5–√ (AC = 10x)
∴ Height of the pole = 100 5–√ m
Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405)
Answer:
Let the height of the peak be “h” mile. Let AD be x mile.
∴ AB = (x + 1) mile.
In the right ∆ ADC, tan 8° = ACAC
0.1405 = hx
x = h0.1405 ………..(1)
In ∆ ABC, tan 4° = ACAB
0.0699 = hx+1 ⇒ (x + 1) 0.0699 = h
0.0699x + 0.0699 = h
0.0699 x = h – 0.0699
x = h−0.06990.0699 ………(2)
Equation (1) and (2) we get,
h−0.06990.0699 = h0.1405
0.0699 h = 0.1405 (h – 0.0699)
0.0699 h = 0.1405 h – 0.0098
0.0098 = 0.1405 h – 0.0699 h
0.0098 = 0.0706 h
h = 0.00980.0706 = 98706 = 0.1388
= 0.14 mile (approximately)
Height of the peak = 0.14 mile
