Unit 6: Introduction to Trigonometry
Review Exercise 6 Solutions
Full solutions to the comprehensive review exercise including MCQs, angle conversions, trigonometric identity proofs, ratio computation, and word problems.
Question 1
MCQs — 10 Questions
Choose the correct option for each of the following:
i. The value of tan−1(12) radians is:
ii. In a right triangle, hypotenuse = 13, angle = 30°. Length of opposite side?
Opposite = 13 × sin 30° = 13 × 1/2 = 6.5
iii. Person 50 m away, elevation angle 45°. Height of building?
h = 50 × tan 45° = 50 × 1 = 50 m
iv. sec2 θ − tan2 θ = ?
From the Pythagorean identity: 1 + tan2θ = sec2θ
v. If sin θ = 3/5, and θ is acute, then cos2 θ = ?
cos2θ = 1 − sin2θ = 1 − 9/25 = 16/25
vi. 5π24 radians = ? degrees
(5π/24) × (180/π) = 5 × 7.5 = 37.5°
vii. 13π8 radians = ? degrees
(13π/8) × (180/π) = 13 × 22.5 = 292.5°
viii. Which is a valid identity?
ix. sin 60° = ?
x. cos2 100° + sin2 100° = ?
Fundamental identity: sin2θ + cos2θ = 1 for all θ
Answer Key
i→d, ii→a, iii→a, iv→b, v→c, vi→b, vii→d, viii→a, ix→d, x→a
Question 2
Angle Conversions
Convert the given angles: (a) degrees to radians in terms of π (b) radians to degrees and minutes.
Solution
(a) Degrees to Radians
Formula: radians = degrees × π180
(i) 255°
= 255 × π180
= 255π180 = 17π12 rad
= 255 × π180
= 255π180 = 17π12 rad
(ii) 75° 45′
75° 45′ = 75 + 4560 = 75.75°
= 75.75 × π180
= 303π720 = 101π240 rad
75° 45′ = 75 + 4560 = 75.75°
= 75.75 × π180
= 303π720 = 101π240 rad
(iii) 142.5°
= 142.5 × π180
= 142.5π180 = 19π24 rad
= 142.5 × π180
= 142.5π180 = 19π24 rad
(b) Radians to Degrees and Minutes
Formula: degrees = radians × 180π
(i) 19π24 rad
= 19π24 × 180π
= 19 × 18024 = 142.5°
= 142° 30′
= 19π24 × 180π
= 19 × 18024 = 142.5°
= 142° 30′
(ii) π12 rad
= π12 × 180π
= 18012 = 15°
= 15° 00′
= π12 × 180π
= 18012 = 15°
= 15° 00′
(iii) 11π16 rad
= 11 × 18016 = 123.75°
0.75° = 0.75 × 60′ = 45′
= 123° 45′
= 11 × 18016 = 123.75°
0.75° = 0.75 × 60′ = 45′
= 123° 45′
Question 3
Trigonometric Identity Proofs
Prove the following trigonometric identities:
Solution
(i) sin θ1 − cos θ = 1 + cos θsin θ
L.H.S = sin θ1 − cos θ
Multiply numerator and denominator by (1 + cos θ):
= sin θ(1 + cos θ)(1 − cos θ)(1 + cos θ)
= sin θ(1 + cos θ)1 − cos2 θ
= sin θ(1 + cos θ)sin2 θ [since 1 − cos2 θ = sin2 θ]
= 1 + cos θsin θ = R.H.S
Multiply numerator and denominator by (1 + cos θ):
= sin θ(1 + cos θ)(1 − cos θ)(1 + cos θ)
= sin θ(1 + cos θ)1 − cos2 θ
= sin θ(1 + cos θ)sin2 θ [since 1 − cos2 θ = sin2 θ]
= 1 + cos θsin θ = R.H.S
Hence Proved.
(ii) sin θ(cosec θ − sin θ) = 1sec2 θ
L.H.S = sin θ(cosec θ − sin θ)
= sin θ × cosec θ − sin θ × sin θ
= 1 − sin2 θ [since sinθ × cosecθ = 1]
= cos2 θ
= 1sec2 θ = R.H.S
= sin θ × cosec θ − sin θ × sin θ
= 1 − sin2 θ [since sinθ × cosecθ = 1]
= cos2 θ
= 1sec2 θ = R.H.S
Hence Proved.
(iii) cosec θ − sec θcosec θ + sec θ = 1 − tan θ1 + tan θ
L.H.S = cosec θ − sec θcosec θ + sec θ = 1sin θ − 1cos θ1sin θ + 1cos θ
Multiply numerator and denominator by sin θ:
= 1 − sin θcos θ1 + sin θcos θ = 1 − tan θ1 + tan θ = R.H.S
Multiply numerator and denominator by sin θ:
= 1 − sin θcos θ1 + sin θcos θ = 1 − tan θ1 + tan θ = R.H.S
Hence Proved.
(iv) tan θ + cot θ = 1sin θ cos θ
L.H.S = tan θ + cot θ
= sin θcos θ + cos θsin θ
= sin2 θ + cos2 θsin θ cos θ
= 1sin θ cos θ = R.H.S
= sin θcos θ + cos θsin θ
= sin2 θ + cos2 θsin θ cos θ
= 1sin θ cos θ = R.H.S
Hence Proved.
(v) cos θ + sin θcos θ − sin θ + cos θ − sin θcos θ + sin θ = 21 − 2 sin2 θ
L.H.S = (cos θ + sin θ)2 + (cos θ − sin θ)2(cos θ − sin θ)(cos θ + sin θ)
Numerator: (a+b)2 + (a−b)2 = 2(a2 + b2) = 2(cos2θ + sin2θ) = 2(1) = 2
Denominator: (a−b)(a+b) = a2 − b2 = cos2θ − sin2θ = (1 − sin2θ) − sin2θ = 1 − 2 sin2θ
L.H.S = 21 − 2 sin2 θ = R.H.S
Numerator: (a+b)2 + (a−b)2 = 2(a2 + b2) = 2(cos2θ + sin2θ) = 2(1) = 2
Denominator: (a−b)(a+b) = a2 − b2 = cos2θ − sin2θ = (1 − sin2θ) − sin2θ = 1 − 2 sin2θ
L.H.S = 21 − 2 sin2 θ = R.H.S
Hence Proved.
(vi) 1 + cos θ1 − cos θ = (cosec θ + cot θ)2
L.H.S = 1 + cos θ1 − cos θ
Divide both numerator and denominator by sin θ:
= 1sin θ + cos θsin θ1sin θ − cos θsin θ
= cosec θ + cot θcosec θ − cot θ
Multiply numerator and denominator by (cosec θ + cot θ):
= (cosec θ + cot θ)2cosec2 θ − cot2 θ
= (cosec θ + cot θ)21 [since 1 + cot2θ = cosec2θ]
= (cosec θ + cot θ)2 = R.H.S
Divide both numerator and denominator by sin θ:
= 1sin θ + cos θsin θ1sin θ − cos θsin θ
= cosec θ + cot θcosec θ − cot θ
Multiply numerator and denominator by (cosec θ + cot θ):
= (cosec θ + cot θ)2cosec2 θ − cot2 θ
= (cosec θ + cot θ)21 [since 1 + cot2θ = cosec2θ]
= (cosec θ + cot θ)2 = R.H.S
Hence Proved.
Question 4
Remaining Ratios
If tan θ = 3√2, find the remaining trigonometric ratios when θ lies in the first quadrant.
Solution
From tan θ = 3√2, we have: perpendicular a = 3 and base c = √2.
By Pythagoras' Theorem: b2 = a2 + c2
b2 = 32 + (√2)2 = 9 + 2 = 11
b = √11
Since θ is in the first quadrant, all ratios are positive:
By Pythagoras' Theorem: b2 = a2 + c2
b2 = 32 + (√2)2 = 9 + 2 = 11
b = √11
Since θ is in the first quadrant, all ratios are positive:
sin θ = 3√11
cos θ = √2√11
cosec θ = √113
sec θ = √11√2
cot θ = √23
Question 5
Word Problem — Elevation
From a point on the ground, the angle of elevation to the top of a 30-metre-high building is 28°. How far is the point from the base of the building?
Solution
Height BC = 30 m, Angle of elevation = 28°, Distance AB = x = ?
tan 28° = mBCmAB = 30x
x = 30tan 28° = 300.5317 = 56.42 m
tan 28° = mBCmAB = 30x
x = 30tan 28° = 300.5317 = 56.42 m
✓ Distance from building = 56.42 m
Question 6
Word Problem — Ladder
A ladder leaning against a wall forms an angle of 65° with the ground. If the ladder is 10 m long, how high does it reach on the wall?
Solution
Ladder length AC = 10 m, Angle with ground = 65°, Height BC = h = ?
sin 65° = mBCmAC = h10
h = 10 × sin 65°
h = 10 × 0.9063 = 9.06 m
sin 65° = mBCmAC = h10
h = 10 × sin 65°
h = 10 × 0.9063 = 9.06 m
✓ Height reached on wall = 9.06 m
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