The length of the chord joining the points $(4\cos{ heta,\ 4\sin heta})$ and $(4\cos( heta + 60{^\circ}),\ 4\sin{( heta + 60{^\circ})})$of the circle $x^{2} + y^{2} = 16$ is

Question

The length of the chord joining the points $(4\cos{	heta,\ 4\sin	heta})$ and $(4\cos(	heta + 60{^\circ}),\ 4\sin{(	heta + 60{^\circ})})$of the circle $x^{2} + y^{2} = 16$ is

Dr. Abubacker · Accepted Answer

(a)
Length of the chord
$= \sqrt{\left\lbrack 4\cos(	heta + 60{^\circ}) - 4\cos	heta ightbrack^{2} + \left\lbrack 4\sin(	heta + 60{^\circ}) - 4\sin	heta ightbrack^{2}}$
$= 4\sqrt{\begin{array}{r}
\cos^{2}{(	heta + 60{^\circ}) + \cos^{2}	heta +} \
\sin^{2}{(	heta + 60{^\circ}) + \sin^{2}{	heta - 2\cos(	heta + 60{^\circ})}}
\end{array}}$
$\cos	heta - 2\sin(	heta + 60{^\circ})\sin	heta$
$= 4\sqrt{1 + 1 - 2\cos{60{^\circ}}} = 4$

Q. The length of the chord joining the points $(4\cos{\theta,\ 4\sin\theta})$ and $(4\cos(\theta + 60{^\circ}),\ 4\sin{(\theta + 60{^\circ})})$of the circle $x^{2} + y^{2} = 16$ is

Explanation

Reviewed by Dr. Abubacker

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