Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.4 Pdf
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TN Samacheer Kalvi 10th Maths Solutions Chapter 6 – Trigonometry
Board | TNSCERT Class 10th Maths |
Class | 10 |
Subject | Maths |
Chapter | 6 (Exercise 6.4) |
Chapter Name | Trigonometry |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise – 6.4
(1) From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree. (√3 = 1.732)
Solution:
Given that, AE = 13m
Now, ∆ABE
tan30° = AE/AB
1/√3 = 13/AB
AB = 13√3
Now, AB = EC = 13√3 m
Now, ∆ECD,
tan45° = DC/EC
1 = DC/13√3
DC = 13√3
Now, AE = BC = 13m
Second tree height
= BD
= BC + CD
= (13 + 13√3)
= 13 + 13 × 1.732 [∵ √3 = 1.732]
= 13 + 22.52
= 35.52 m
Thus, the second tree height is 35.52 m.
(2) A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill. (√3 = 1.732)
Solution:
Given that, AE = 40m.
Now, ∆ABE,
tan30° = AE/AB
1/√3 = 40/AB
AB = 40√3
AB = 40×1.732 (Given √3 = 1.732)
AB = 69.28m
Now, AB = EC = 69.28m
Now, ∆ECD
tan60° = DC/EC
√3 = DC/40√3
DC = 120m
Now, the distance of the hill from the ship is = AC = 69.28m
And the height of the hill is = BD
= BC + CD
= (40 + 120) [∵ DC = AE]
= 160m
(3) If the angle of elevation of a cloud from a point ‘h’ metres above a lake isθ1 and the angle of depression of its reflection in the lake is θ2. Prove that the height that the cloud is located from the ground is h(tanθ1 + tanθ2)/tanθ2 – tanθ1
Solution:
Let, AB = the surface of the lake given that AE = BC = h m
Let, BD = the height of the cloud is located from the ground = x m.
CD = BD – BC
= (x – h) m
Now, ∆ECD,
tanθ1 = DC/EC
tanθ1 = (x-h)/EC —- (i)
Now, BF is the reflection of the BD
∴ BF = B = x m
Now, ∆EFC,
tanθ2 = Fc/EC
tanθ2 = x+h/EC —– (ii)
Now, (i) + (ii), we get,
tanθ1 + tanθ2 = x-h/EC + x+h/EC
tanθ1 + tanθ2 = x-h+x+h/EC
tanθ1 + tanθ2 = 2x/EC
EC = 2x/tanθ1 + tanθ2 —– (iii)
Now, (ii) – (i), we get,
tanθ2 – tanθ1 = x+h/EC – x-h/EC
tanθ2 – tanθ1 = x+h-x+h/EC
tanθ2 – tanθ1 = 2h/EC
EC = 2h/tanθ2 – tanθ1
2x/tanθ1 + tanθ2 = 2h/tanθ2 – tanθ1 [By equation (iii)]
x/tanθ1 + tanθ2 = h/tanθ2 – tanθ1
x = h(tanθ1 + tanθ2)/tanθ2 – tanθ1
Thus, the height of the cloud is located from the ground is h(tanθ1 + tanθ2)/tanθ2 – tanθ1 [Proved]
(4) The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30°. If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms.
Solution:
AE = apartment height = 50 m.
Now, AE = BC = 50m
∆ABE,
tan30° = AE/AB
1/√3 = 50/AB
AB = 50√3
and ∆ABD,
tan60° = BD/AB
√3 = BD/50√3
BD = 50×3
BD = 150m
BD = the height of the cell phone tower.
∴ BD = 150m
Therefore, the height of the cell phone tower is 150m.
Now, 150m > 120m
Yes, the height of the above mentioned tower meet the radiation norms.
(5) The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find
(i) The height of the lamp post.
(ii) The difference between height of the lamp post and the apartment.
(iii) The distance between the lamp post and the apartment. (√3 = 1.732)
Solution:
AE = The lamp post
BC = The apartment = 66m
Now, ∆ABC
tan30° = BC/AB
1/√3 = 66/AB
AB = 66√3
Now, AB = DC = 66√3
∆DCE,
tan60° = DE/DC
√3 = DE/66√3
DE = 198m
(i) The height of the lamp post is = AE
= AD + DE
= (66 + 198) m
= 264 m
(ii) The difference between height of the lamp post and apartment is = DE = 198m
(iii) The distance between the lamp post and the apartment is = AB = 66√3 m
= 66 × 1.732 = 114.31 m
(6) Three villagers A, B and C can see each other using telescope across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30°. Calculate:
(i) The vertical height between A and B
(ii) The vertical between B and C. (tan20° = 0.3640, √3 = 1.732)
Solution:
Now, BE = 8km and BD = 12km.
Now, ∆BEA,
tan20° = AE/EB
0.3640 = AE/8 [Give tan20° = 0.3640]
AE = 2.912 m
Now, ∆BDC,
tan30° = DC/BD
1/√3 = DC/12
DC = 4√3
DC = 4×1.732 (Given √3 = 1.732)
DC = 6.92m
(i) The vertical height between A and B is 2.91 km.
(ii) The vertical height between B and C is 6.92m
Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 6 Trigonometry Exercise 6.4
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