Q 1. In ΔABC, right-angled at B, AB = 24cm, BC = 7cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Given:
In ΔABC
∠B = 90°
AB = 24
BC = 7
To Find:
(i) sin A =?
Cos A =?
(ii) sin C =?
Cos C =?
Solution:
By PGT
(AC)2 = (AB)2+(BC )2
=> (AC)2 = (24 )2 + (7 )2
=> (AC)2 = 576 + 49
=> (AC)2 = 625
=> AC = √625
=> AC = 25
(i) For ∠A
P = 24
b = 7
H = 25
(i) sin A = P/H = 24/25
cos A = B/H = 7/25
(ii) For ∠C
P = 7
b = 24
H = 25
sin C = 7/25
cos C = b/H = 24/25
Q 2. In Fig. 8.13,
find: tan P – cot R
(PR)2 = (PQ)2 + (QR)2
=> (13)2 = (12 )2 + (QR)2
=> 169 = 144 + (QR)2
=> 169 – 144 = (QR)2
=> 25 = (QR)2
=> √25 = QR
=> 5 = QR
For ∠P
P = 5
b = 12
H = 13
tanP = P/H = 5/13
For √ R
P = 12
b = 5
H = 13
cot R = b/H = 5/13
tanP-cotR
= 5/13 – 5/13
= 0
Q3. If sin A = 3/4 , calculate cos A and tan A.
solution:
sin A = 3/4 = P/H
P = 3
H = 4
H = √H2 – P2
=> H = √42– 32
=> H = √16 – 9
=> H = √7
cos A = b/H = 4/√7
tan A = P/b = 3/√7
Q4. Given 15 cot A = 8, Find sin A and sec A
Solution:
15 cot A = 8
cot A = 8/15 = b/P
b = 8
P = 15
H =?
By PGT
H = √P2+b2
=> H = √(15)2 + (8)2
=> H = √225 + 64
=> H = √289
=> H = √17×17
=> H = 17
sin A = P/H = 15/17
sec A = b/H = 8/17
Q5. If sec θ = 13/12, calculate all other trigonometric ratios.
Solution:
sec A = 13/12 = H/b
H = 13
b = 12
P = √H2 – b2
=> P = √132 – 122
=>P = √132 – 122
=>P = √169 – 144
=>P = √25
P = 5
For ∠ θ
P = 5
b = 12
H = 13
sin θ = P/H = 5/13
cos θ = b/H = 12/13
tan θ = P/b = 5/12
cot θ = b/P = 12/5
sec θ = H/b = 13/12
cosec θ = H/P = 13/5
Q6. If ∠A and ∠B are acute angles such that cos A=cos B, then show that ∠A = ∠B.
Given:
∠A < 90º and
∠B < 90°
cos A = cos B
To Show:
∠A = ∠B
Solution:
In ΔABC
∠C = 90°
For ∠A
P = BC
b = AC
H = AB
Cos A = b/H
=> Cos A = AC/AB …………………..(i)
For ∠B
P = AC
b = BC
H = AB
Cos B = b/H
=>Cos B = BC/AB …………………..(ii)
Cos A = Cos B [ Given ]
AC/AB = BC/AB [ From Equation(i) and Equation(ii) ]
=> AC = BC
∴ ΔABC is an isosceles triangle
We know that in isosceles triangle equal sides of opposite angles are also equal
∴ ∠A = ∠B
Q 7. If cot θ = 7/8, evaluate:
(i) (1 + sin θ)(1 – sin θ)/(1 + cos θ ) (1 – cos θ)
(ii) cot2 θ
Solution:
cot θ = 7/8 = b/P
b = 7
p = 8
H =?
By PGT
H = √P2 + b2
=> H = √82 + 72
=> H = √64 + 49
=> H = √113
sin θ = P/H = 8/√113
cos θ = b/H = 7/√113
(i) (1+sin θ) (1 – sin θ) / (1 + cos θ ) (1 – cos θ)
=> (1 + 8/√113) (1 – 8/√113) / (1 + 7/√113) (1 – 7/√113)
=> [(1)2-(8/√113)2 ] / [ (1)2 – (7/√113)2 ]
=> [ 1-64/113] / [ 1 – 49 /113 ]
=> [(113 – 64)/113] / [ ( 113 – 49/113) ]
=> (47 / 113) / ( 64/113)
=> 47/64 ans
(ii) cot2θ = (7/8)2 = 49/64
Q8). If 3 cot A = 4, check whether 1 – tan2 A / 1 + tan2A = cos2 A – sin2 A or not.
Given:
3 cot A = 4
To Check:
1 – tan2 A / 1 + tan2A = cos2 A – sin2 A or not
Solution:
3 cot A = 4
=> cot A = 4/3 = b/P
b = 4
P = 3
H =?
H = √32 + 42
=> H = √ (9 + 16)
=> H = √25
=> H = 5
sin A = P/H = 3/5
cos A = b/H = 4/5
tan A = P/B = 3/4
1 – tan2 A / 1 + tan2 A = cos2 A – sin2A
L.H.S. = 1 – tan2A / 1 + tan2A
= 1 – (3/4)2 / 1 + (3/4)2
= [1 – 9/16 ]/ [1 + 9/16]
= [ (16 – 9) /16 ] / [ (16 + 9) /16 ]
= [ (7)/16] / [ (25)/16 ]
= 7/25
R.H.S. = cos2 A – sin2A
=> (4/5)2 – (3/5)2
=> 16/25 – 9/25
=> (16-9) / 25
=> 7/25
L.H.S. = R.H.S. ( Hance Proved )
Q.9) In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of:
(i)sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Given:
In Δ ABC
∠B = 90º
tan A = 1/√3
To Find:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
tan A = 1/√3 = P/H
P = 1
b = √3
H2 = P 2 + b 2
H2 = (1 )2 + (√3)2
=> H2 = (1 )2 + (√3)2
=> H2 = 1 + 3
=> H2 = 4
=> H = 2
For ∠A
P = 1
b = √3
H = 2
sin A = P/H = 1/2
cos A = b/H = √3/2
For ∠C
P = √3
b = 1
H = 2
sin C = P/H = √3/2
cos C = b/H = 1/2
(i) sin A cos C + cos A sin C
= 1/2×1/2 + √3/2 × √3/2
= 1/4 + 3/4
= 4/4
= 1
(ii) cos A cos C – sin A sin C
= √3/2×1/2 – 1/2 × √3/2
= √3/4 – √3/4
= 0
Q.10) In ΔPQR, right-angled at Q, PR + QR = 25cm and PQ = 5cm. Determine the values of sin P, Cos P, and tan P.
Given:
In ΔPQR
∠Q = 90º
PR + QR = 25cm
PQ = 5cm
To Find:
sin p =?
cos p =?
tan p =?
Solution:
According to figure
P = 5cm
PR + QR = 25cm
=>H+b = 25
=>H = 25-b ……………(i)
H2 = p2+b2
=> (25-b)2 = 52+b2
=> (25)2 – 2×25×b + b2 = 25 + b2
=> 625 – 50b + b2 = 25 + b2
=> – 50b + b2 – b2 = 25 – 625
=> – 50b = – 600
=> b = – 600/- 50
=> b = 12 Put into equation number (i)
H = 25 – b ……………(i)
=> H = 25 – 12
=>H = 13
now For ∠p
p = 5
b = 12
H = 13
sin P = P/H = 5/13
cos P = b/H = 12/13
tan P = P/b = 5/12
Q.11) State whether the following are true or false. justify your answer.
(i) The value of tan A is always less than 1.
Ans: False
justification:
(ii) sec A = 12/5 for some value of angle A.
Ans: True
(iii) cos A is the abbreviation used for the cosecant of angle A.
Ans: True
(iv) cot A is the product of cot and A
Ans: False
(v) sin θ = 4/3 for some angle θ.
Ans: False