ABC is a right-angled isosceles triangle with | vedclass

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Question

(A) In $	riangle ABC$,$\angle B = 90^\circ$ and it is isosceles,so $\angle CAB = \angle ACB = 45^\circ$.Given $\angle DCB = 15^\circ$,then $\angle AC

Vedclass · Accepted Answer

$35\sqrt{2} \, cm$

Vedclass · Answer

(A) In $	riangle ABC$,$\angle B = 90^\circ$ and it is isosceles,so $\angle CAB = \angle ACB = 45^\circ$.Given $\angle DCB = 15^\circ$,then $\angle ACD = \angle ACB - \angle DCB = 45^\circ - 15^\circ = 30^\circ$.In $	riangle DBC$,$\angle B = 90^\circ$ and $\angle DCB = 15^\circ$,so $\angle BDC = 75^\circ$.Let $BC = h$. Since $	riangle ABC$ is isosceles,$AB = BC = h$.Then $BD = AB - AD = h - 35$.In $	riangle DBC$,$	an(15^\circ) = \frac{BD}{BC} = \frac{h - 35}{h}$.We know $	an(15^\circ) = 2 - \sqrt{3}$.$h(2 - \sqrt{3}) = h - 35 \implies h(1 - \sqrt{3}) = -35 \implies h = \frac{35}{\sqrt{3} - 1} = \frac{35(\sqrt{3} + 1)}{2}$.Now,in $	riangle DBC$,$\cos(15^\circ) = \frac{BC}{CD} \implies CD = \frac{BC}{\cos(15^\circ)} = \frac{h}{\cos(15^\circ)}$.Using $\cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$,we get $CD = \frac{35(\sqrt{3} + 1)}{2} 	imes \frac{4}{\sqrt{6} + \sqrt{2}} = \frac{70(\sqrt{3} + 1)}{\sqrt{2}(\sqrt{3} + 1)} = \frac{70}{\sqrt{2}} = 35\sqrt{2} \, cm$.

$ABC$ is a right-angled isosceles triangle with $\angle B = 90^\circ$. If $D$ is a point on $AB$ such that $\angle DCB = 15^\circ$ and $AD = 35 \, cm$,then $CD = $

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