Chapter 6: Problem 15
(a) \(\sin \theta=1\). (b) \(\theta=\frac{\pi}{2}\)
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Chapter 6: Problem 15
(a) \(\sin \theta=1\). (b) \(\theta=\frac{\pi}{2}\)
These are the key concepts you need to understand to accurately answer the question.
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An arc \(\mathrm{PQ}\) subtends an angle of \(60^{\circ}\) at the centre of a circle of radius \(1 \mathrm{~cm}\). The length of \(\mathrm{PQ}\) is: (a) \(60 \mathrm{~cm}\) (b) \(30 \mathrm{~cm}\) (c) \(\frac{\pi}{6} \mathrm{~cm}\) (d) \(\frac{\pi}{3} \mathrm{~cm}\) (e) \(\frac{\pi^{2}}{18} \mathrm{~cm}\).
\(f(\theta) \equiv \cos \theta\) (a) For \(-\frac{\pi}{2}<\theta<\frac{\pi}{2}, f(\theta)>0\) (b) \(f(\theta)\) is undefined when \(\theta=(2 n+1) \frac{\pi}{2}\). (c) \(-1 \leqslant f(\theta) \leqslant 1\). (d) \(f(\theta)\) is periodic with a period of \(\pi\).
\(\mathrm{P}\) and \(\mathrm{Q}\) are points on a circle of radius \(r\), and the chord PQ subtends an angle \(2 \theta\) radians at its centre 0 . If \(A\) is the area enclosed by the minor arc \(\mathrm{PQ}\) and the chord \(\mathrm{PQ}\), and if \(B\) is the area enclosed by the arc \(\mathrm{PQ}\) and the tangents to the circle at \(\mathrm{P}\) and \(\mathrm{Q}\), prove that $$ A-B \equiv r^{2}(2 \theta-\tan \theta-\sin \theta \cos \theta) $$
The associated acute angle for \(280^{\circ}\) is: (a) \(100^{\circ}\) (b) \(10^{\circ}\) (c) \(80^{\circ}\) (d) \(-80^{\circ}\) (e) \(190^{\circ}\).
\(\sin \theta=0 \quad\) when \(\theta=n \pi\)
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