Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.2 Pdf
TN Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.2: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.2 – Trigonometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 6 Exercise 6.2 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 6.2
TN Samacheer Kalvi 10th Maths Solutions Chapter 6 – Trigonometry
| Board | TNSCERT Class 10th Maths |
| Class | 10 |
| Subject | Maths |
| Chapter | 6 (Exercise 6.2) |
| Chapter Name | Trigonometry |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise – 6.2
(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.
Solution:
Let AB be the tower and the angle elevation from c is θ.
Given that AB = 10√3 m and BC = 30m
Now, ∆ABC
AB/BC = tanθ
tanθ = 10√3/30
tanθ = √3/3
tanθ = 1/√3
tanθ = tan30°
θ = 30°
Therefore, the required angles is 30°.
(2) A road is flanked on either side by continuous rows of houses of height 4√3 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.
Solution:
Let DB = x
Given that DC = BC = x/2
∴ BC = x/2
Given that, AB = 4√3
Now, ∆ABC,
tanθ = AB/BC
tan30° = 4√3/(x/2) [∵θ = 30°]
1/√3 = 4√3/(x/2)
x/2 = 4×3 = 12
x = 24
∴ BD = width of the road therefore 24m width of the road.
(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)
Solution:
Now, CD = window
Let, CD = hm and AC = xm
Now, ∆ABC,
tanθ = AC/AB
tan45° = x/5 [∵θ = 45°]
1 = x/5
x = 5
∴ Now, AD = AC + CD = x + CD = (5 + h) m
Now, ∆ABD
tanθ = AD/AB
tan60° = 5+h/5 [∵θ = 60°]
√3 = 5+h/5
5+h = 5√3
h = 5√3 – 5
h = 5×1.732 – 5 [∵ √3 = 1.732]
h = 8.660 – 5
h = 3.66m
Thus, the window height is 3.66m.
(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. (tan40° = 0.8391, √3 = 1.732)
Solution:
CD = statue and AC = pedestal.
Given that CD = 1.6m.
Let AC = h m and AB = xm
AD = AC + CD = (h + 1.6) m
∆ABC,
tan 40° = AC/AB
0.8391 = h/x [∵ tan40° = 0.8391]
x = h/0.8391 —- (i)
Now, ∆ABD,
tan60° = AD/AB
√3 = h+1.6/x
x = h + 1.6/√3
h/0.8391 = h+1.6/1.732 [By equation (i) and √3 = 1.732]
1.732h = 0.8391h + 1.6× 0.8391
1.732h – 0.8391h = 1.6 × 0.8391
0.8929h = 1.6 × 0.8391
h = 1.6×0.8391/0.8929
h = 1.6×8391/8929
h = 1.6×0.93
h = 1.5m
Thus, the pedestal height is 1.5m.
(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)
Solution:
Now, AF = AD = radius = r
Now, AB = AF + FB = (r + 7) m.
Now, AC = AF + FB + BC = r + 7 + 5 = (r + 12) m.
and AE = AD + DE = (r + h) m.
∆ABE
tan45° = AE/AB
1 = r + h/r + 7
r + 7 = r + h
h = 7m
∆ACD,
tan30° = AD/AC
1/√3 = r/r+12
√3r = r + 12
(√3 – 1) r = 12
r = 12/√3 – 1
r = 12/1.732 – 1
r = 12/.732
r = 16.39m
Therefore, the height of the pole is 7m. and radius of the dome is 16.39 m.
(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?
Solution:
Given that AB = 15m
Let, BE = elective pole height = h
Let, BC = x m
AE = AB – BE = (15 – h) m
Now, ∆ABC,
tan60° = AB/BC
√3 = 15/x
x = 15/√3
x = 5√3 m
Now, ∆AED, tan30° = AE/ED
1/√3 = 15-h/5√3
15 – h = 5
h = 15 – 5 = 10m
Thus, the height of the elective pole is 40m.
Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 6 Trigonometry Exercise 6.2
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.