Samacheer Kalvi 10th Maths Solutions Chapter 6 Unit Exercise 6 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 (Unit Exercise 6) |
Chapter Name | Trigonometry |
TNSCERT Class 10th Maths Pdf | all Unit Exercise Solution
Unit Exercise – 6
(1) Prove that
(i) cot2 A (secA-1/1+sinA) + sec2A (sinA-1/1+secA) = 0
(ii) tan2θ – 1/tan2θ + 1 = 1 – 2 cos2θ
Solution:
(i) cot2A (secA – 1/1+sinA) + sec2A (sinA-1/1+secA)
= cot2A (secA-1)/1+sinA + sec2A(sinA-1)/1+secA
= 1-1/(1+sinA) (1+secA)
= 0/(1+sinA) (1+secA)
= 0
Thus, cot2A (secA-1/1+sinA) + sec2A (sinA-1/1+secA) = 0 [Proved]
(ii) tan2θ – 1/tan2θ + 1
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= cos2θ (sin2θ – cos2θ)/cos2θ (sin2θ + cos2 θ)
= 1 – cos2θ – cos2θ/1 [∵ sin2θ + cos2θ = 1]
= 1 – 2 cos2θ
Thus, tan2 θ – 1/tan2 θ + 1 = 1 – 2 cos2 θ [Proved]
(2) Prove that (1+sinθ – cosθ/1+sinθ + cosθ)2 = 1- cosθ/1+cosθ
Solution:
Now, (1 + sinθ – cosθ/1+sinθ+cosθ)2
= {(1+sinθ)-cosθ/(1+sinθ)+cosθ}2
= 1 (1+sinθ) – cosθ (1-sinθ)/1 (1+sinθ)+cosθ(1+sinθ)
= (1+sinθ) (1-cosθ)/(1+sinθ) (1+cosθ)
= 1-cosθ/1+cosθ [Proved]
(3) If x sin3θ + y cos3θ = sinθ.cosθ and x sinθ = ycosθ, then prove that x2 + y2 = 1.
Solution:
Given that x sinθ = ycos θ —- (i)
∴x sin3 θ + y cos3 θ = sinθcosθ
x sin3θ + y cosθ.cos2θ = sinθcosθ
x sin3θ + x sinθ cos2θ = sinθ.cosθ [By —– (i)]
xsinθ (sin2θ + cos2θ) = sinθ.cosθ
xsinθ = sinθ.cosθ [∵ sin2θ + cos2θ = 1]
x = cosθ —- (ii)
∴xsinθ = ycosθ
Cosθ sinθ = ycosθ [By equation (ii)]
y = sinθ
∴ x = cosθ and y = sinθ
∴Now, x2 + y2
= cos2θ + sin2θ
= 1
Thus, x2 + y2 = 1 [Proved]
(4) If a cosθ – b sinθ = c, then prove that (a sinθ + b cosθ) = ±√a2 + b2 – c2
Solution:
Given that a cosθ – bsinθ = c
Both sides squaring, we get
(acosθ – b sinθ)2 = c2
a2 cos2θ – b2sin2θ = c2
a2 (1 – sin2θ) + b2 (1 – cos2θ) 2ab cosθsinθ = c2
a2 – a2sin2 + b2 – b2 cos2θ 2ab cosθsinθ = c2
a2 + b2 – a2 sin2θ – b2cos2θ – 2ab cosθsinθ = c2
a2 + b2 – (a2 sin2θ + 2ab sinθ.cosθ + b2 cos2θ) = c2
a2 + b2 – (a sinθ + b cosθ)2 = c2
(asinθ + b cosθ)2 = a2 + b2 – c2
asinθ + bcosθ = ± √a2 + b2 – c2
Thus, (a sinθ + bcosθ) = ± √a2 + b2 – c2 [Proved]
(5) A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevationof the bird from the same point is 30°. Determine the speed at which the bird flies. (√3 = 1.732)
Solution:
Now, BC = DE = ground bird height = 80m.
Now, ∆ABC,
tan45° = BC/AB
1 = 80/AB
AB = 80m
tan30° = DE/AE
1/√3 = 80/AE
AE = 80√3
AE = 80×1.732 (∵ √3 = 1.732)
AE = 138.56m
Now, The bird fly the distance at 2 seconds is = DC = BE = AE – AB
= (138.56 – 80) m
= 58.56 m
Thus, the bird fly speed is = 58.56/2 m/s
= 29.28 m/s
(6) An aeroplane is flying parallel to the Earth’s surface at a speed of 175 m/sec and at a height of 600 m. The angle of elevation of the aeroplane from a point on the Earth’s surface is 37°. After what period of time does the angle of elevation increase to 53°? (tan53° = 1.3270, tan37° = 0.7536)
Solution:
Given BC = DE = 600m
Let, t times period the angle of elevation increase to 53°.
So, distance for CD = BE = 175t [∵ speed = 175 m/s]
Now, AE = AB + BE
= AB + 175t
Now, ∆ABC,
tan53° = CB/AB
1.327° = 600/AB
AB = 600/1.3270 —- (i)
Now, ∆AED,
tan37° = DE/AE
0.7536 = 600/AB + 175t
AB + 175t = 600/0.7536
AB = 600/0.7536 – 175t
600/1.3270 = 600/0.7536 – 175t [By equation (i)]
175t = 600/0.7536 – 600/1.3270
175t = 600 (1/0.7536 – 1/1.3270)
175t = 600×0.57
t = 600×0.57/175
t = 344.03/175
t = 1.96 second
Thus, the required time taken 1.96 second
(7) A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away.
(i) How far is B to the North of A?
(ii) How far is B to the West of A?
(iii) How far is C to the North of B?
(iv) How far is C to the East of B?
(sin55° = 0.8192, cos55° = 0.573, sin42° = 0.6691, cos42° = 0.7431)
Solution:
Now, AB = 30km and BC = 32km
(i) For is 13 to the North of A is = BD
∆ABD,
sin55° = BD/AB
0.8192 = BD/30
BD = 0.8192 × 30
BD = 24.57 km
Thus, 24.57 km for B to the North of A.
(ii) For is B to the West of A is = AD
Now, ∆ABD,
Cos55° = AD/AB
0.5736 = AD/30
AD = 17.2 km
Thus, 17.2 km for is B to the west of A.
(iii) For is C to the North of B is = EC
Now, ∆BEC
Sin42° = EC/BC
0.6691 = EC/32
EC = 21.41 km
Thus, 21.41 km for is C to the North of B.
(iv) For 13 e to the East of B is = BE
∆BEC,
Cos42° = BE/BC
0.7431 = BE/32
BE = 23.77 km
Thus, 23.77 km for is C to the East of B.
(8) Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45°respectively. If the distance between the ships is 200 (√3+1/√3) meters, find the height of the lighthouse.
Solution:
Let, AC = lighthouse height = x m,
Given, BD = 200 (√3+1/√3) m.
Now, ∆ABC
tan60° = AC/BC
√3 = x/BC
BC = x/√3 —– (i)
and ∆ACD,
tan45° = AC/CD
1 = x/CD
CD = x —– (ii)
Now, BD = 200 (√3+1/√3)
BC + CD = 200 (√3+1/√3)
x/√3 + x = 200 (√3+1/√3) [By equation (i) and (ii)]
x (1/v3 + 1) = 200 (1 + 1/√3)
x = 200 m
Thus, the lighthouse height is 200 meters.
(9) A building and a statue are in opposite side of a street from each other 35 m apart. From a point on the roof of building the angle of elevation of the top of statue is 24° and the angle of depression of base of the statue is 34°. Find the height of the statue.
(tan24° = 0.4452, tan34° = 0.6745)
Solution:
Given that BC = AD = 35m
Now, AB = CD
Let, EC = x m
Now, ∆ABC
tan34° = AB/BC
0.6745 = AB/35
∴ AB = 23.60
Now, DE = EC – DC
= EC – AB
= B (x – 23.60) m.
Now, ∆ADE,
tan24° = DE/AD
0.4452 = x-23.60/35
x – 23.60 = 15.58
x = 15.58 + 23.60
x = 39.18 m
Thus, the height of the state is 39.18 meters.
Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 6 Trigonometry Unit Exercise 6
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