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A. \( \frac{1003\pi}{72000} \) रेडियन/\( \frac{1003\pi}{72000} \) radians
Step 1
Concept
\(2^\circ30'45''=\frac{1003}{400}^\circ\) and the radian value is \( \frac{1003\pi}{72000} \). First convert to degrees and multiply by \( \frac{\pi}{180} \).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{1003\pi}{72000} \) रेडियन / \( \frac{1003\pi}{72000} \) radians. \(2^\circ30'45''=\frac{1003}{400}^\circ\) and the radian value is \( \frac{1003\pi}{72000} \). First convert to degrees and multiply by \( \frac{\pi}{180} \).
Step 3
Exam Tip
\(2^\circ30'45''=\frac{1003}{400}^\circ\) और रेडियन \( \frac{1003\pi}{72000} \) है। पहले डिग्री में बदलकर \( \frac{\pi}{180} \) से गुणा करें।
\( \frac{37\pi}{15}\times\frac{180^\circ}{\pi}=444^\circ\). If the denominator is (15), use \(180^\circ\div15=12^\circ\).
Step 2
Why this answer is correct
The correct answer is B. \(444^\circ\). \( \frac{37\pi}{15}\times\frac{180^\circ}{\pi}=444^\circ\). If the denominator is (15), use \(180^\circ\div15=12^\circ\).
Step 3
Exam Tip
\( \frac{37\pi}{15}\times\frac{180^\circ}{\pi}=444^\circ\) है। हर (15) हो तो \(180^\circ\div15=12^\circ\) लें।
\( -\frac{49\pi}{12}+\frac{60\pi}{12}=\frac{11\pi}{12} \). Add enough multiples of \(2\pi\) to a negative radian angle.
Step 2
Why this answer is correct
The correct answer is C. \( \frac{11\pi}{12} \). \( -\frac{49\pi}{12}+\frac{60\pi}{12}=\frac{11\pi}{12} \). Add enough multiples of \(2\pi\) to a negative radian angle.
Step 3
Exam Tip
\( -\frac{49\pi}{12}+\frac{60\pi}{12}=\frac{11\pi}{12} \) है। ऋणात्मक रेडियन कोण में \(2\pi\) के पर्याप्त गुणज जोड़ें।
\( \frac{31\pi}{8}-\frac{16\pi}{8}=\frac{15\pi}{8} \) and the reference angle is \(2\pi-\frac{15\pi}{8}=\frac{\pi}{8}\). First find the principal angle.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{\pi}{8} \). \( \frac{31\pi}{8}-\frac{16\pi}{8}=\frac{15\pi}{8} \) and the reference angle is \(2\pi-\frac{15\pi}{8}=\frac{\pi}{8}\). First find the principal angle.
Step 3
Exam Tip
\( \frac{31\pi}{8}-\frac{16\pi}{8}=\frac{15\pi}{8} \) और संदर्भ कोण \(2\pi-\frac{15\pi}{8}=\frac{\pi}{8}\) है। पहले मुख्य कोण निकालें।
\( \frac{19\pi}{7}-2\pi=\frac{5\pi}{7} \), and \( \frac{5\pi}{7} \) lies in the second quadrant. Always reduce the angle before deciding the quadrant.
Step 2
Why this answer is correct
The correct answer is C. तृतीय चतुर्थांश / Third quadrant. \( \frac{19\pi}{7}-2\pi=\frac{5\pi}{7} \), and \( \frac{5\pi}{7} \) lies in the second quadrant. Always reduce the angle before deciding the quadrant.
Step 3
Exam Tip
\( \frac{19\pi}{7}-2\pi=\frac{5\pi}{7} \) है और यह \( \frac{\pi}{2} \) तथा \( \pi \) के बीच है? सावधानी से \( \frac{5\pi}{7} \) द्वितीय चतुर्थांश में है।
\(1480^\circ-1440^\circ=40^\circ\), and it lies in the first quadrant. First find the principal angle for large angles.
Step 2
Why this answer is correct
The correct answer is C. \(40^\circ\). \(1480^\circ-1440^\circ=40^\circ\), and it lies in the first quadrant. First find the principal angle for large angles.
Step 3
Exam Tip
\(1480^\circ-1440^\circ=40^\circ\) है और यह प्रथम चतुर्थांश में है। बड़े कोण में पहले मुख्य कोण निकालें।
\( \theta=\frac{s}{r}=\frac{15\pi}{18}=\frac{5\pi}{6}=150^\circ \). First find the radian angle and then convert to degrees.
Step 2
Why this answer is correct
The correct answer is C. \(150^\circ\). \( \theta=\frac{s}{r}=\frac{15\pi}{18}=\frac{5\pi}{6}=150^\circ \). First find the radian angle and then convert to degrees.
Step 3
Exam Tip
\( \theta=\frac{s}{r}=\frac{15\pi}{18}=\frac{5\pi}{6}=150^\circ \) है। पहले रेडियन कोण निकालकर डिग्री में बदलें।
From \( \frac{81\pi}{8}=\frac{1}{2}\times81\times\theta \), \( \theta=\frac{\pi}{4}=45^\circ \). The angle from the sector area formula is in radians.
Step 2
Why this answer is correct
The correct answer is B. \(45^\circ\). From \( \frac{81\pi}{8}=\frac{1}{2}\times81\times\theta \), \( \theta=\frac{\pi}{4}=45^\circ \). The angle from the sector area formula is in radians.
Step 3
Exam Tip
\( \frac{81\pi}{8}=\frac{1}{2}\times81\times\theta \) से \( \theta=\frac{\pi}{4}=45^\circ \) है। क्षेत्रफल सूत्र में कोण रेडियन में निकलता है।
Using \(A=\frac{1}{2}rs\), \(121=\frac{1}{2}r\cdot22\), so (r=11) and \( \theta=\frac{s}{r}=2 \). When arc and area are given, find (r) first.
Step 2
Why this answer is correct
The correct answer is B. (2) रेडियन / (2) radians. Using \(A=\frac{1}{2}rs\), \(121=\frac{1}{2}r\cdot22\), so (r=11) and \( \theta=\frac{s}{r}=2 \). When arc and area are given, find (r) first.
Step 3
Exam Tip
\(A=\frac{1}{2}rs\) से \(121=\frac{1}{2}r\cdot22\) इसलिए (r=11) और \( \theta=\frac{s}{r}=2 \) है। चाप और क्षेत्रफल साथ दिए हों तो पहले (r) निकालें।
C. \( \frac{3\pi}{5} \) रेडियन/\( \frac{3\pi}{5} \) radians
Step 1
Concept
The minute hand turns \(2\pi\) in (60) minutes, so in (18) minutes it turns \( \frac{18}{60}\cdot2\pi=\frac{3\pi}{5} \). Use the time fraction of one full revolution.
Step 2
Why this answer is correct
The correct answer is C. \( \frac{3\pi}{5} \) रेडियन / \( \frac{3\pi}{5} \) radians. The minute hand turns \(2\pi\) in (60) minutes, so in (18) minutes it turns \( \frac{18}{60}\cdot2\pi=\frac{3\pi}{5} \). Use the time fraction of one full revolution.
Step 3
Exam Tip
मिनट सुई (60) मिनट में \(2\pi\) घूमती है इसलिए (18) मिनट में \( \frac{18}{60}\cdot2\pi=\frac{3\pi}{5} \) है। समय का अनुपात पूर्ण चक्कर से लगाएं।
At (4:40), the minute hand is at \(240^\circ\) and the hour hand is at \(140^\circ\), so the difference is \(100^\circ\). The hour hand moves \(0.5^\circ\) per minute.
Step 2
Why this answer is correct
The correct answer is C. \(100^\circ\). At (4:40), the minute hand is at \(240^\circ\) and the hour hand is at \(140^\circ\), so the difference is \(100^\circ\). The hour hand moves \(0.5^\circ\) per minute.
Step 3
Exam Tip
(4:40) पर मिनट सुई \(240^\circ\) और घंटे सुई \(140^\circ\) पर होती है इसलिए अंतर \(100^\circ\) है। घंटे सुई हर मिनट \(0.5^\circ\) चलती है।
For coterminal angles the difference can be \(360^\circ\), so \(5x+20^\circ=x+380^\circ\) gives \(x=90^\circ\). Form a linear equation.
Step 2
Why this answer is correct
The correct answer is D. \(90^\circ\). For coterminal angles the difference can be \(360^\circ\), so \(5x+20^\circ=x+380^\circ\) gives \(x=90^\circ\). Form a linear equation.
Step 3
Exam Tip
सहसमापी होने पर अंतर \(360^\circ\) हो सकता है इसलिए \(5x+20^\circ=x+380^\circ\) से \(x=90^\circ\) है। रैखिक समीकरण बनाएं।
(3\theta+45^\circ-\(\theta-135^\circ\)=360^\circ) gives \(2\theta+180^\circ=360^\circ\) and \( \theta=90^\circ \). Keep the difference as a multiple of \(360^\circ\).
Step 2
Why this answer is correct
The correct answer is A. \(90^\circ\). (3\theta+45^\circ-\(\theta-135^\circ\)=360^\circ) gives \(2\theta+180^\circ=360^\circ\) and \( \theta=90^\circ \). Keep the difference as a multiple of \(360^\circ\).
Step 3
Exam Tip
(3\theta+45^\circ-\(\theta-135^\circ\)=360^\circ) से \(2\theta+180^\circ=360^\circ\) और \( \theta=90^\circ \) है। अंतर को \(360^\circ\) का गुणज रखें।
Supplementary angles sum to \( \pi \), so \( \pi-\frac{7\pi}{12}=\frac{5\pi}{12} \). In radians use \( \pi \) instead of \(180^\circ\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{5\pi}{12} \). Supplementary angles sum to \( \pi \), so \( \pi-\frac{7\pi}{12}=\frac{5\pi}{12} \). In radians use \( \pi \) instead of \(180^\circ\).
Step 3
Exam Tip
पूरक कोणों का योग \( \pi \) होता है इसलिए \( \pi-\frac{7\pi}{12}=\frac{5\pi}{12} \) है। रेडियन में \(180^\circ\) की जगह \( \pi \) लें।
Complementary angles sum to \( \frac{\pi}{2} \), so \( \frac{\pi}{2}-\frac{5\pi}{18}=\frac{2\pi}{9} \). Use a common denominator before subtracting.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{2\pi}{9} \). Complementary angles sum to \( \frac{\pi}{2} \), so \( \frac{\pi}{2}-\frac{5\pi}{18}=\frac{2\pi}{9} \). Use a common denominator before subtracting.
Step 3
Exam Tip
संपूरक कोणों का योग \( \frac{\pi}{2} \) होता है इसलिए \( \frac{\pi}{2}-\frac{5\pi}{18}=\frac{2\pi}{9} \) है। समान हर बनाकर घटाएं।
\( -\frac{23\pi}{10}+\frac{40\pi}{10}=\frac{17\pi}{10} \), which lies between \( \frac{3\pi}{2} \) and \(2\pi\). First find the principal angle.
Step 2
Why this answer is correct
The correct answer is D. \( \frac{3\pi}{2}<\theta<2\pi \). \( -\frac{23\pi}{10}+\frac{40\pi}{10}=\frac{17\pi}{10} \), which lies between \( \frac{3\pi}{2} \) and \(2\pi\). First find the principal angle.
Step 3
Exam Tip
\( -\frac{23\pi}{10}+\frac{40\pi}{10}=\frac{17\pi}{10} \) है जो \( \frac{3\pi}{2} \) और \(2\pi\) के बीच है। पहले मुख्य कोण निकालें।
A. \( \frac{5\pi}{12} \) और तृतीय चतुर्थांश/\( \frac{5\pi}{12} \) and third quadrant
Step 1
Concept
\( \frac{17\pi}{12} \) lies between \( \pi \) and \( \frac{3\pi}{2} \), and the reference angle is \( \frac{17\pi}{12}-\pi=\frac{5\pi}{12} \). The formula changes by quadrant.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{5\pi}{12} \) और तृतीय चतुर्थांश / \( \frac{5\pi}{12} \) and third quadrant. \( \frac{17\pi}{12} \) lies between \( \pi \) and \( \frac{3\pi}{2} \), and the reference angle is \( \frac{17\pi}{12}-\pi=\frac{5\pi}{12} \). The formula changes by quadrant.
Step 3
Exam Tip
\( \frac{17\pi}{12} \) \( \pi \) और \( \frac{3\pi}{2} \) के बीच है और संदर्भ कोण \( \frac{17\pi}{12}-\pi=\frac{5\pi}{12} \) है। चतुर्थांश के अनुसार सूत्र बदलता है।
\( \frac{5}{3}\times\frac{180^\circ}{3}=100^\circ \). Use the approximation of \( \pi \) given in the question.
Step 2
Why this answer is correct
The correct answer is C. \(100^\circ\). \( \frac{5}{3}\times\frac{180^\circ}{3}=100^\circ \). Use the approximation of \( \pi \) given in the question.
Step 3
Exam Tip
\( \frac{5}{3}\times\frac{180^\circ}{3}=100^\circ \) है। दिए गए \( \pi \) के अनुमान का ही उपयोग करें।
The rate is (6) radians per second, so in (3) seconds it turns (18) radians, which is \(18\cdot\frac{180^\circ}{\pi}=\frac{3240^\circ}{\pi}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{1080^\circ}{\pi} \). The rate is (6) radians per second, so in (3) seconds it turns (18) radians, which is \(18\cdot\frac{180^\circ}{\pi}=\frac{3240^\circ}{\pi}\).
Step 3
Exam Tip
दर (6) रेडियन प्रति सेकंड है इसलिए (3) सेकंड में (18) रेडियन यानी \(18\cdot\frac{180^\circ}{\pi}=\frac{3240^\circ}{\pi}\) है।
\(225^\circ=\frac{5\pi}{4}\) and \(r=\frac{s}{\theta}=\frac{25\pi}{5\pi/4}=20\). Convert the angle to radians first.
Step 2
Why this answer is correct
The correct answer is B. (20) सेमी / (20) cm. \(225^\circ=\frac{5\pi}{4}\) and \(r=\frac{s}{\theta}=\frac{25\pi}{5\pi/4}=20\). Convert the angle to radians first.
Step 3
Exam Tip
\(225^\circ=\frac{5\pi}{4}\) और \(r=\frac{s}{\theta}=\frac{25\pi}{5\pi/4}=20\) है। कोण को पहले रेडियन में बदलें।
\(144^\circ=\frac{4\pi}{5}\) and \(40\pi=\frac{1}{2}r^2\cdot\frac{4\pi}{5}\) gives (r=10). Keep the angle in radians in the area formula.
Step 2
Why this answer is correct
The correct answer is C. (10) सेमी / (10) cm. \(144^\circ=\frac{4\pi}{5}\) and \(40\pi=\frac{1}{2}r^2\cdot\frac{4\pi}{5}\) gives (r=10). Keep the angle in radians in the area formula.
Step 3
Exam Tip
\(144^\circ=\frac{4\pi}{5}\) और \(40\pi=\frac{1}{2}r^2\cdot\frac{4\pi}{5}\) से (r=10) है। क्षेत्रफल सूत्र में कोण रेडियन में रखें।
\(2\theta=\frac{22\pi}{9}\) and \( \frac{22\pi}{9}-2\pi=\frac{4\pi}{9} \). After multiplying, subtract \(2\pi\).
Step 2
Why this answer is correct
The correct answer is B. \( \frac{4\pi}{9} \). \(2\theta=\frac{22\pi}{9}\) and \( \frac{22\pi}{9}-2\pi=\frac{4\pi}{9} \). After multiplying, subtract \(2\pi\).
Step 3
Exam Tip
\(2\theta=\frac{22\pi}{9}\) और \( \frac{22\pi}{9}-2\pi=\frac{4\pi}{9} \) है। गुणा करने के बाद \(2\pi\) घटाएं।
\( \alpha+\beta=\frac{20\pi}{10}=2\pi \), which is coterminal with (0). In principal angle form, \(2\pi\) is taken as (0).
Step 2
Why this answer is correct
The correct answer is A. (0). \( \alpha+\beta=\frac{20\pi}{10}=2\pi \), which is coterminal with (0). In principal angle form, \(2\pi\) is taken as (0).
Step 3
Exam Tip
\( \alpha+\beta=\frac{20\pi}{10}=2\pi \) है जो (0) के साथ सहसमापी है। मुख्य कोण में \(2\pi\) को (0) माना जाता है।
\(875^\circ-720^\circ=155^\circ\), and the reference angle is \(180^\circ-155^\circ=25^\circ\). Find the principal angle before the reference angle.
Step 2
Why this answer is correct
The correct answer is A. \(25^\circ\). \(875^\circ-720^\circ=155^\circ\), and the reference angle is \(180^\circ-155^\circ=25^\circ\). Find the principal angle before the reference angle.
Step 3
Exam Tip
\(875^\circ-720^\circ=155^\circ\) और संदर्भ कोण \(180^\circ-155^\circ=25^\circ\) है। मुख्य कोण के बाद संदर्भ कोण निकालें।
\( -\theta=-\frac{5\pi}{4} \), and \( -\frac{5\pi}{4}+2\pi=\frac{3\pi}{4} \). Add \(2\pi\) to a negative angle.
Step 2
Why this answer is correct
The correct answer is B. \( \frac{3\pi}{4} \). \( -\theta=-\frac{5\pi}{4} \), and \( -\frac{5\pi}{4}+2\pi=\frac{3\pi}{4} \). Add \(2\pi\) to a negative angle.
Step 3
Exam Tip
\( -\theta=-\frac{5\pi}{4} \) और \( -\frac{5\pi}{4}+2\pi=\frac{3\pi}{4} \) है। ऋणात्मक कोण में \(2\pi\) जोड़ें।
\( \theta-\pi=\frac{14\pi}{5}-\frac{5\pi}{5}=\frac{9\pi}{5} \), and it is in (0) to \(2\pi\). First subtract and then check the range.
Step 2
Why this answer is correct
The correct answer is D. \( \frac{9\pi}{5} \). \( \theta-\pi=\frac{14\pi}{5}-\frac{5\pi}{5}=\frac{9\pi}{5} \), and it is in (0) to \(2\pi\). First subtract and then check the range.
Step 3
Exam Tip
\( \theta-\pi=\frac{14\pi}{5}-\frac{5\pi}{5}=\frac{9\pi}{5} \) है और यह (0) से \(2\pi\) में है। पहले घटाव करें फिर सीमा देखें।
\(3\theta=\frac{15\pi}{6}=\frac{5\pi}{2}\), and its principal angle is \( \frac{\pi}{2} \). Hence the terminal side lies on the positive (y)-axis.
Step 2
Why this answer is correct
The correct answer is D. ऋणात्मक (y)-अक्ष / Negative (y)-axis. \(3\theta=\frac{15\pi}{6}=\frac{5\pi}{2}\), and its principal angle is \( \frac{\pi}{2} \). Hence the terminal side lies on the positive (y)-axis.
Step 3
Exam Tip
\(3\theta=\frac{15\pi}{6}=\frac{5\pi}{2}\) और इसका मुख्य कोण \( \frac{\pi}{2} \) है। इसलिए अंतिम भुजा धनात्मक (y)-अक्ष पर है।
\( \frac{7\pi}{18}\times\frac{180^\circ}{\pi}=70^\circ \). Multiply by \( \frac{180^\circ}{\pi} \) to convert radians to degrees.
Step 2
Why this answer is correct
The correct answer is A. \(70^\circ\). \( \frac{7\pi}{18}\times\frac{180^\circ}{\pi}=70^\circ \). Multiply by \( \frac{180^\circ}{\pi} \) to convert radians to degrees.
Step 3
Exam Tip
\( \frac{7\pi}{18}\times\frac{180^\circ}{\pi}=70^\circ \) है। रेडियन से डिग्री में \( \frac{180^\circ}{\pi} \) से गुणा करें।
B. \( \frac{23\pi}{24} \) रेडियन/\( \frac{23\pi}{24} \) radians
Step 1
Concept
\(137^\circ30'=137.5^\circ\), and \(137.5^\circ\times\frac{\pi}{180}=\frac{55\pi}{72}\).
Step 2
Why this answer is correct
The correct answer is B. \( \frac{23\pi}{24} \) रेडियन / \( \frac{23\pi}{24} \) radians. \(137^\circ30'=137.5^\circ\), and \(137.5^\circ\times\frac{\pi}{180}=\frac{55\pi}{72}\).
Step 3
Exam Tip
\(137^\circ30'=137.5^\circ\) और \(137.5^\circ\times\frac{\pi}{180}=\frac{55\pi}{72}\) है।
In the third quadrant the principal angle is \(180^\circ+35^\circ=215^\circ\). Add or subtract the reference angle according to the quadrant.
Step 2
Why this answer is correct
The correct answer is B. \(215^\circ\). In the third quadrant the principal angle is \(180^\circ+35^\circ=215^\circ\). Add or subtract the reference angle according to the quadrant.
Step 3
Exam Tip
तीसरे चतुर्थांश में मुख्य कोण \(180^\circ+35^\circ=215^\circ\) होगा। चतुर्थांश के अनुसार संदर्भ कोण जोड़ें या घटाएं।
In the fourth quadrant \( \theta=2\pi-\frac{\pi}{9}=\frac{17\pi}{9} \). A full revolution in radians is \(2\pi\).
Step 2
Why this answer is correct
The correct answer is C. \( \frac{17\pi}{9} \). In the fourth quadrant \( \theta=2\pi-\frac{\pi}{9}=\frac{17\pi}{9} \). A full revolution in radians is \(2\pi\).
Step 3
Exam Tip
चतुर्थ चतुर्थांश में \( \theta=2\pi-\frac{\pi}{9}=\frac{17\pi}{9} \) है। रेडियन में पूर्ण चक्कर \(2\pi\) होता है।
\( \frac{7\pi}{9}=140^\circ \), which is greater than \(90^\circ\), so no positive complementary angle is possible. First make the units same.
Step 2
Why this answer is correct
The correct answer is D. कोई धनात्मक मान नहीं / No positive value. \( \frac{7\pi}{9}=140^\circ \), which is greater than \(90^\circ\), so no positive complementary angle is possible. First make the units same.
Step 3
Exam Tip
\( \frac{7\pi}{9}=140^\circ \) है जो \(90^\circ\) से बड़ा है इसलिए धनात्मक संपूरक कोण संभव नहीं है। पहले इकाई समान करें।
\( \frac{11\pi}{6}=330^\circ \), and the other angle is \(330^\circ-275^\circ=55^\circ\). First make the units of both angles the same.
Step 2
Why this answer is correct
The correct answer is C. \(55^\circ\). \( \frac{11\pi}{6}=330^\circ \), and the other angle is \(330^\circ-275^\circ=55^\circ\). First make the units of both angles the same.
Step 3
Exam Tip
\( \frac{11\pi}{6}=330^\circ \) और दूसरा कोण \(330^\circ-275^\circ=55^\circ\) है। पहले दोनों कोणों की इकाई समान करें।
\( -\frac{13\pi}{6}+\frac{24\pi}{6}=\frac{11\pi}{6} \), and the reference angle is \( \frac{\pi}{6} \). First bring the negative angle into (0) to \(2\pi\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{\pi}{6} \). \( -\frac{13\pi}{6}+\frac{24\pi}{6}=\frac{11\pi}{6} \), and the reference angle is \( \frac{\pi}{6} \). First bring the negative angle into (0) to \(2\pi\).
Step 3
Exam Tip
\( -\frac{13\pi}{6}+\frac{24\pi}{6}=\frac{11\pi}{6} \) और संदर्भ कोण \( \frac{\pi}{6} \) है। ऋणात्मक कोण को पहले (0) से \(2\pi\) में लाएं।
\( \frac{4\pi}{3}+3\pi=\frac{13\pi}{3} \), and \( \frac{13\pi}{3}-4\pi=\frac{\pi}{3} \). Subtract multiples of \(2\pi\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{\pi}{3} \). \( \frac{4\pi}{3}+3\pi=\frac{13\pi}{3} \), and \( \frac{13\pi}{3}-4\pi=\frac{\pi}{3} \). Subtract multiples of \(2\pi\).
Step 3
Exam Tip
\( \frac{4\pi}{3}+3\pi=\frac{13\pi}{3} \) और \( \frac{13\pi}{3}-4\pi=\frac{\pi}{3} \) है। \(2\pi\) के गुणज घटाएं।
Sector area is \( \frac{1}{2}rs=\frac{1}{2}\times6\times9=27 \). Use this short formula when arc length is given.
Step 2
Why this answer is correct
The correct answer is B. (27) वर्ग सेमी / (27) square cm. Sector area is \( \frac{1}{2}rs=\frac{1}{2}\times6\times9=27 \). Use this short formula when arc length is given.
Step 3
Exam Tip
त्रिज्यखंड क्षेत्रफल \( \frac{1}{2}rs=\frac{1}{2}\times6\times9=27 \) है। चाप लंबाई दी हो तो यह छोटा सूत्र लगाएं।
The difference \( \frac{29\pi}{12}-\frac{5\pi}{12}=2\pi \), so they are coterminal. In radians, coterminal angles differ by a multiple of \(2\pi\).
Step 2
Why this answer is correct
The correct answer is A. वे सहसमापी हैं / They are coterminal. The difference \( \frac{29\pi}{12}-\frac{5\pi}{12}=2\pi \), so they are coterminal. In radians, coterminal angles differ by a multiple of \(2\pi\).
Step 3
Exam Tip
अंतर \( \frac{29\pi}{12}-\frac{5\pi}{12}=2\pi \) है इसलिए वे सहसमापी हैं। रेडियन में सहसमापी अंतर \(2\pi\) का गुणज होता है।
The supplementary angle is \(180^\circ-112^\circ30'=67^\circ30'\). Be careful with borrowing in minute subtraction.
Step 2
Why this answer is correct
The correct answer is A. \(67^\circ30'\). The supplementary angle is \(180^\circ-112^\circ30'=67^\circ30'\). Be careful with borrowing in minute subtraction.
Step 3
Exam Tip
पूरक कोण \(180^\circ-112^\circ30'=67^\circ30'\) है। मिनट वाले घटाव में उधार लेने का ध्यान रखें।
The complementary angle is \(90^\circ-38^\circ45'=51^\circ15'\). Treat \(90^\circ\) as \(89^\circ60'\) while subtracting.
Step 2
Why this answer is correct
The correct answer is A. \(51^\circ15'\). The complementary angle is \(90^\circ-38^\circ45'=51^\circ15'\). Treat \(90^\circ\) as \(89^\circ60'\) while subtracting.
Step 3
Exam Tip
संपूरक कोण \(90^\circ-38^\circ45'=51^\circ15'\) है। \(90^\circ\) को \(89^\circ60'\) मानकर घटाएं।
