Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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C. \(\frac{13\pi}{18}\)
दूसरे चतुर्थांश में मुख्य कोण \(\pi-\alpha\) होता है। इसलिए \(\pi-\frac{5\pi}{18}=\frac{13\pi}{18}\) है। / In the second quadrant, the principal angle is \(\pi-\alpha\). Hence \(\pi-\frac{5\pi}{18}=\frac{13\pi}{18}\).
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If the principal angle of \(-\theta\) is \(110^\circ\), what is the principal angle of \(\theta\)?
D. \(250^\circ\)
\(-\theta\) का मुख्य कोण \(110^\circ\) है, इसलिए \(\theta\) का मुख्य कोण \(-110^\circ+360^\circ=250^\circ\) होगा। परीक्षा में संकेत बदलने पर \(360^\circ\) से समायोजन करें। / The principal angle of \(-\theta\) is \(110^\circ\), so the principal angle of \(\theta\) is \(-110^\circ+360^\circ=250^\circ\). In exams, adjust with \(360^\circ\) after changing the sign.
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B. (4)
एक पूर्ण चक्कर \(360^\circ\) होता है और \(\frac{1440^\circ}{360^\circ}=4\)। परीक्षा में अंतर को \(360^\circ\) से भाग दें। / One complete revolution is \(360^\circ\) and \(\frac{1440^\circ}{360^\circ}=4\). In exams, divide the difference by \(360^\circ\).
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What is the decimal degree measure of \(73^\circ20'24''\)?
A. \(73.34^\circ\)
\(20'=\frac{1}{3}^\circ\) और \(24''=\frac{1}{150}^\circ\), इसलिए कुल \(73.34^\circ\) है। परीक्षा में मिनट और सेकंड को अलग-अलग डिग्री में बदलें। / \(20'=\frac{1}{3}^\circ\) and \(24''=\frac{1}{150}^\circ\), so the total is \(73.34^\circ\). In exams, convert minutes and seconds separately into degrees.
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In which quadrant will the terminal side of \(\frac{107\pi}{12}\) lie?
C. दूसरा चतुर्थांशSecond quadrant
\(\frac{107\pi}{12}\) में से \(8\pi\) घटाने पर \(\frac{11\pi}{12}\) मिलता है। \(\frac{11\pi}{12}\) दूसरा चतुर्थांश में है। / Subtracting \(8\pi\) from \(\frac{107\pi}{12}\) gives \(\frac{11\pi}{12}\). \(\frac{11\pi}{12}\) lies in the second quadrant.
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D. \(\frac{\pi}{6}\)
छोटा प्रतिघड़ी कोण \(2\pi-\frac{11\pi}{6}=\frac{\pi}{6}\) है। परीक्षा में एक पूर्ण चक्कर से बड़ा भाग घटाएं। / The smaller counterclockwise angle is \(2\pi-\frac{11\pi}{6}=\frac{\pi}{6}\). In exams, subtract the larger part from one full rotation.
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A sector has radius (7) cm and perimeter (22) cm. What is the central angle in radians?
B. \(\frac{8}{7}\)
सेक्टर परिमाप \(2r+r\theta\) होता है, इसलिए \(14+7\theta=22\)। इससे \(\theta=\frac{8}{7}\) मिलता है। / Sector perimeter is \(2r+r\theta\), so \(14+7\theta=22\). This gives \(\theta=\frac{8}{7}\).
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If a sector has arc length (16) cm and central angle (4) radians, what is its area?
A. (32) वर्ग सेमी(32) square cm
\(s=r\theta\) से (r=4), फिर \(A=\frac{1}{2}r^2\theta=32\)। परीक्षा में पहले त्रिज्या निकालकर क्षेत्रफल लगाएं। / From \(s=r\theta\), (r=4), then \(A=\frac{1}{2}r^2\theta=32\). In exams, first find the radius and then apply the area formula.
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If \(126^\circ=\frac{m\pi}{20}\) radians, what is the value of (m)?
C. (14)
\(126^\circ=\frac{126\pi}{180}=\frac{7\pi}{10}=\frac{14\pi}{20}\), इसलिए (m=14)। परीक्षा में दिए रूप जैसा हर बनाएं। / \(126^\circ=\frac{126\pi}{180}=\frac{7\pi}{10}=\frac{14\pi}{20}\), so (m=14). In exams, make the denominator match the given form.
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If \(300^\circ=\frac{p\pi}{9}\) radians, what is the value of (p)?
D. (15)
\(300^\circ=\frac{5\pi}{3}=\frac{15\pi}{9}\), इसलिए (p=15)। परीक्षा में डिग्री से रेडियन बदलकर समान हर बनाएं। / \(300^\circ=\frac{5\pi}{3}=\frac{15\pi}{9}\), so (p=15). In exams, convert degrees to radians and then match denominators.
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