IPMPrepFind the minimum value of 15 sin θ − 20 cos θ + 5.
Solution
15 sin θ − 20 cos θ has minimum −√(225 + 400) = −25.
Minimum = 5 − 25 = −20.
Answer: (D) −20
Quantitative Ability · Geometry
Trigonometry: IPMAT practice questions
24 questions, 7 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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Find sin2 18° + sin2 36° + sin2 54° + … + sin2 90°.
Solution
sin2 x + sin2(90° − x) = sin2 x + cos2 x = 1, so pair the terms from both ends.
There are 2 such pairs, plus sin2 90° = 1.
Total = 3.
Answer: 3
If tan A + cot A = 2, find tan10 A + cot10 A.
Solution
t + 1/t = 2 forces t = 1 (since (t − 1)2 = 0).
So tan A = cot A = 1 and the value is 1 + 1 = 2.
Answer: 2
Evaluate 2 tan2 30° + 2 cos2 60° + 3 cot2 30°.
Solution
Use the standard values: tan2 30° = 13; cos2 60° = 14; cot2 30° = 3.
Value = 616.
Answer: (A) 616
Evaluate 3 sin2 90° + 2 cot2 30° − 4 sec2 45°.
Solution
Use the standard values: sin2 90° = 1; cot2 30° = 3; sec2 45° = 2.
Value = 1.
Answer: (C) 1
Find the minimum value of 20 sin θ − 21 cos θ + 16.
Solution
20 sin θ − 21 cos θ has minimum −√(400 + 441) = −29.
Minimum = 16 − 29 = −13.
Answer: −13
If sin θ = 35 and θ is acute, find tan θ + sec θ.
Solution
cos θ = √(1 − 925) = 45.
tan θ = 34, sec θ = 54; sum = 2.
Answer: (C) 2
From a point on the ground, the angle of elevation of the top of a 20 m tower is 30°. How far is the point from the foot of the tower?
Solution
tan 30° = 20/d, so d = 20/tan 30°.
tan 30° = 1/√3, so d = 20√3 m.
Answer: (B) 20√3 m
From a point on the ground, the angle of elevation of the top of a 30 m tower is 60°. How far is the point from the foot of the tower?
Solution
tan 60° = 30/d, so d = 30/tan 60°.
tan 60° = √3, so d = 10√3 m.
Answer: (D) 10√3 m
If tan A + cot A = 2, find tan20 A + cot20 A.
Solution
t + 1/t = 2 forces t = 1 (since (t − 1)2 = 0).
So tan A = cot A = 1 and the value is 1 + 1 = 2.
Answer: 2
Find sin2 15° + sin2 30° + sin2 45° + … + sin2 90°.
Solution
sin2 x + sin2(90° − x) = sin2 x + cos2 x = 1, so pair the terms from both ends.
There are 2 such pairs, plus sin2 45° = 12, plus sin2 90° = 1.
Total = 3.5.
Answer: 3.5
Find the maximum value of 15 sin θ + 20 cos θ.
Solution
a sin θ + b cos θ ranges between −√(a2 + b2) and √(a2 + b2).
Maximum = √(225 + 400) = 25.
Answer: (C) 25
If cos θ = 45 and θ is acute, find cosec θ − cot θ.
Solution
sin θ = 35.
cosec θ = 53, cot θ = 43; difference = 13.
Answer: (D) 13
Find sin2 30° + sin2 60° + sin2 90°.
Solution
sin2 x + sin2(90° − x) = sin2 x + cos2 x = 1, so pair the terms from both ends.
There are 1 such pairs, plus sin2 90° = 1.
Total = 2.
Answer: 2
If tan A + cot A = 2, find tan5 A + cot5 A.
Solution
t + 1/t = 2 forces t = 1 (since (t − 1)2 = 0).
So tan A = cot A = 1 and the value is 1 + 1 = 2.
Answer: (C) 2
If sin θ + cos θ = 43, find sin θ cos θ.
Solution
Square both sides: 1 + 2 sin θ cos θ = 169.
sin θ cos θ = (169 − 1)/2 = 718.
Answer: (D) 718
If sin θ = 817 and θ is acute, find tan θ + sec θ.
Solution
cos θ = √(1 − 64289) = 1517.
tan θ = 815, sec θ = 1715; sum = 53.
Answer: (B) 53
The angles of elevation of the top of a tower from two points on the same side, in line with its foot, are 30° and 60°. If the points are 10 m apart, find the height of the tower.
Solution
Let the height be x. Distances from the foot: x√3 (at 30°) and x/√3 (at 60°).
x√3 − x/√3 = 10, so 2x/√3 = 10 and x = 10√32 = 5√3 m.
Answer: (D) 5√3 m
Evaluate 3 sec2 60° − 2 cos2 30° + cos2 60°.
Solution
Use the standard values: sec2 60° = 4; cos2 30° = 34; cos2 60° = 14.
Value = 434.
Answer: (C) 434
If cos θ = 2425 and θ is acute, find cosec θ − cot θ.
Solution
sin θ = 725.
cosec θ = 257, cot θ = 247; difference = 17.
Answer: (B) 17
The angles of elevation of the top of a tower from two points on the same side, in line with its foot, are 30° and 60°. If the points are 75 m apart, find the height of the tower.
Solution
Let the height be x. Distances from the foot: x√3 (at 30°) and x/√3 (at 60°).
x√3 − x/√3 = 75, so 2x/√3 = 75 and x = 75√32 = 75√32 m.
Answer: (A) 75√32 m
Evaluate cos2 0° + 4 sec2 60° + cot2 30°.
Solution
Use the standard values: cos2 0° = 1; sec2 60° = 4; cot2 30° = 3.
Value = 20.
Answer: (A) 20
Find sin2 3° + sin2 6° + sin2 9° + … + sin2 90°.
Solution
sin2 x + sin2(90° − x) = sin2 x + cos2 x = 1, so pair the terms from both ends.
There are 14 such pairs, plus sin2 45° = 12, plus sin2 90° = 1.
Total = 15.5.
Answer: 15.5
Find the maximum value of 8 sin θ + 15 cos θ.
Solution
a sin θ + b cos θ ranges between −√(a2 + b2) and √(a2 + b2).
Maximum = √(64 + 225) = 17.
Answer: (B) 17
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.