JEE Main Sets, Relations, Functions & Trigonometry: Trigonometry – Chapter Test

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Difficulty: 10 medium · 10 hard

Trigonometry MCQ questions with answers

The first 3 of 20 questions. Unlock the series for the rest, each with a solution.

1. Q1. The principal value of cos⁻¹(cos(13π/6)) is:

   cos⁻¹(cos(13π/6)) का मुख्य मान है:

   • A. 13π/6
   • B. -π/6
   • C. π/6
   • D. 11π/6
   Solution

   cos(13π/6) = cos(2π + π/6) = cos(π/6) = √3/2, and since the range of cos⁻¹ is [0, π], the principal value is π/6.

   cos(13π/6) = cos(2π + π/6) = cos(π/6) = √3/2, तथा चूँकि cos⁻¹ का परिसर [0, π] है, अतः मुख्य मान π/6 है।

2. Q2. From the top of a cliff 100 m high, the angles of depression of two boats, lying in the same vertical plane as the observer and on the same side of the cliff, are 45° and 30°. The distance between the two boats is:

   100 मीटर ऊँची एक चट्टान के शिखर से, प्रेक्षक के समान ऊर्ध्वाधर तल में तथा चट्टान के एक ही ओर स्थित दो नावों के अवनमन कोण 45° और 30° हैं। दोनों नावों के बीच की दूरी है:

   • A. 100(√3 + 1) m / 100(√3 + 1) मीटर
   • B. 100(√3 − 1) m / 100(√3 − 1) मीटर
   • C. 50(√3 + 1) m / 50(√3 + 1) मीटर
   • D. 100√3 m / 100√3 मीटर
   Solution

   The horizontal distances of the boats from the foot of the cliff are 100 tan 45°⁻¹-based, i.e. 100 cot 45° = 100 m and 100 cot 30° = 100√3 m, so the separation is 100√3 − 100 = 100(√3 − 1) m.

   चट्टान के पाद से नावों की क्षैतिज दूरियाँ 100 cot 45° = 100 मीटर तथा 100 cot 30° = 100√3 मीटर हैं, अतः दोनों के बीच की दूरी 100√3 − 100 = 100(√3 − 1) मीटर है।

3. Q3. The angle of elevation of the top of a vertical tower from a point on the ground is 30°. On walking 20 m directly towards the base of the tower, the angle of elevation becomes 60°. The height of the tower is:

   भूमि पर स्थित एक बिंदु से एक ऊर्ध्वाधर मीनार के शिखर का उन्नयन कोण 30° है। मीनार के आधार की ओर सीधे 20 m चलने पर उन्नयन कोण 60° हो जाता है। मीनार की ऊँचाई है:

   • A. 10 m
   • B. 20√3 m
   • C. 10√3 m
   • D. 20 m
   Solution

   With h the height, h cot30° − h cot60° = 20, so h(√3 − 1/√3) = 20, giving h(2/√3) = 20 and h = 10√3 m.

   यदि ऊँचाई h हो, तो h cot30° − h cot60° = 20, अतः h(√3 − 1/√3) = 20, जिससे h(2/√3) = 20 तथा h = 10√3 m।

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