Two tangents are drawn from a point P to the | vedclass

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Question

$	an 	heta=\frac{12}{5}$

$PA =\cot \frac{	heta}{2}$

$	herefore$ area of $\Delta PAB =\frac{1}{2}( PA )^{2} \sin 	heta=\frac{1}{2} \cot ^{2}

Vedclass · Accepted Answer

$9:4$

Vedclass · Answer

$	an 	heta=\frac{12}{5}$

$PA =\cot \frac{	heta}{2}$

$	herefore$ area of $\Delta PAB =\frac{1}{2}( PA )^{2} \sin 	heta=\frac{1}{2} \cot ^{2} \frac{	heta}{2} \sin 	heta$

$=\frac{1}{2}\left(\frac{1+\cos 	heta}{1-\cos 	heta}ight) \sin 	heta$

$=\frac{1}{2}\left(\frac{1+\frac{5}{13}}{1-\frac{5}{13}}ight)\left(\frac{12}{13}ight)=\frac{1}{2} \frac{18}{18} 	imes \frac{2}{13}=\frac{27}{26}$

area of $\Delta CAB =\frac{1}{2} \sin 	heta=\frac{1}{2}\left(\frac{12}{13}ight)=\frac{6}{13}$

$	herefore \frac{	ext { area of } \Delta PAB }{	ext { area of } \Delta CAB }=\frac{9}{4}$

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