Telangana SCERT Solution Class X (10) Maths Chapter 11 Trigonometry Exercise 11.2
Exercise 11.2
1> Evaluate the Following
i> Sin450 + Cos 450
= 1/√2 + 1/√2
= 1+1/√2
= 2/√2
= √2 x √2 / √2
= √2
2> Cos 450/sec300 + cosec 600
Solution :
Cos450 / sec300 + Cosec600 = 1/√2/2√3 + 2√3
= 1/√2/2+2√3
= 1/√2/4/√3
= 1/√2 x √3/4
= √3 / 4√2
3> Sin300 + tan450 + Cosec600/Cot450 + Cos600 – sec300
Solution:
Sin300 + tan450 + Cosec600 / Cot450 + cos600 – Sec300
= ½ + 1 -2√3 / 1+1/2 – 2/3
= 1
4> 2tan2450 + Cos230 – Sin2600
Solution:
2tan2450 + Cos230 – Sin2600
= 2(1)2 + (√3/2)2 – (√3/2)2
= 2 + ¾ – ¾
= 2
5> Sec2600 – tan2600/sin2300 + Cos2 300
Solution:
Sec2600 – tan2600 / Sin2300 = (2)2 – (√3)2/(1/2)+( √3/2)2
= 4-3/1/4+3/4
= 4-3/4/4
= 1/1
= 1
2> Choose the right option and Justify your choice
i> 2tan300/1+tan2450
a) sin600 b) Cos600 c) Tan300 d) Sin300
Solution:
2tan300 / 1+tan2450 = 2×1/√3 / 1+ (1)2
= 2 x 1/√3 / 1+1
= 2 x 1/√3 / 2
= 1/√3
2tan300/1-tan2300
a) Cos600 b) Sin600 c) tan600 d) sin300
Solution:
2 tan300/1-tan230 = 2 x 1/√3 / 1-(1/√3)2
= 2/√3/1-1/3
= 2/√3/3-1/3
= 2/√3/2/3
= 3/√3
= √3 x √3 /√3
= √3
= tan 60o
3>
Solution :
Sin60o + Cos30o + sin30o Cos600
√3/2 x √3/2 + ½ x ½
= ¾ + ¼
= 3 + 1 / 4
= 4/4
= 1
Sin (60o + 30o ) = Sin (90o)
= Y
Therefore, We can Conclude that,
Sin (60o + 300) = Sin600Cos300+ Sin300 Cos600
4> Is it right to say that Cos (60o + 30o) = Cos60o Cos300 – Sin60o sin30o
Solution:
L.H.S = Cos (60o + 30o)
= Cos90o
= 0
R.H.S = Cos60o cos30o – Sin600 + sin300
= ½ x √3/2 – √3/2 x ½
= √3/4 – √3/4
= 0
From equation (1) & (2)
Therefore, it is right to say that,
Cos (600 + 300) = Cos 600 Cos300 – Sin600 Sin300
5> In right angled triangle △ PQR, right angle is at Q , PQ = 6cm and ∠ RPQ = 60o , Determine the lengths of QR and PR
Solution:
In Right Triangle PQR,
∠ Q = 90o , PQ = 6cm
and ∠ RPQ = 60o
(Therefore, Given)
tan600 = Opp. Side/ Adj Side
QR/PQ
Tan600 = QR/6
√3 = QR/6
6√3 = QR
QR = 6√3cm
Cos600 = Adj. Side
= PQ/PR
½ = 6/PR
PR = 2 x 6
PR = 12cm
6>
In △ XYZ = YZ/XZ
= X/2X
sin ∠ YXZ = ½ = Sin30o
therefore, ∠ yxz = 30o
Therefore, ∠ yxz = 30o
Cos ∠ YZX = Adj. Side/hypotenuse
= YZ/XZ
= x/2x
= Sin ∠ yxz = yz/xz
= X / 2X
sin ∠ yxz = ½ = sin30o
Cos ∠ yzx = Adj. Side/hypotenuse
= YZ/XZ
= X/2X
Cos ∠ yzx = ½
Cos ∠ YZX = Cos600
∠ YZX = 60o
Therefore ∠ YXZ = 30o and ∠ YZX = 60o
7> Is It right to say that (A+B) = SinA + SinB ? Justify your answer ,
Let A = 300 and B = 60o
L.H.S = Sin (A+B)
= Sin (300 + 60o)
= Sin900
= Y
R.H.S = SinA + SinB
= Sin30o + Sin60o
= ½ + √3/2
= 1+√3/2
From equation (1) & (2)
LHS ≠ R.H.S
Therefore, It is not right to say that Sin (A+B) = Sin A + Sin B
Here is your solution of Telangana SCERT Class 10 Maths Chapter 11 Trigonometry Exercise 11.2
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