In a right triangle ABC, right angled at A, o | vedclass

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Question

$	herefore$ $\frac{{AC}}{{AD}}$ $= \frac{1}{{\sin 	heta }}$

$\Rightarrow$ $\frac{{AD}}{{AB}}$ $=sin \,\left( {\frac{\pi }{2} - 	heta }

Vedclass · Accepted Answer

$\frac{{AB \cdot AD}}{{\sqrt {A{B^2} - A{D^2}} }}$

Vedclass · Answer

$	herefore$ $\frac{{AC}}{{AD}}$ $= \frac{1}{{\sin 	heta }}$

$\Rightarrow$ $\frac{{AD}}{{AB}}$ $=sin \,\left( {\frac{\pi }{2} - 	heta } ight)$ $=cos\,\,	heta$

$\Rightarrow$ $\frac{{A{D^2}}}{{A{C^2}}} + \frac{{A{D^2}}}{{A{B^2}}}$ $=1$

$\Delta = $ $\frac{1}{2}$ $\sqrt {{x^2} + {{m{l}}^2}} \,\cdot\,y$ $=x^2 l^2$

$\Rightarrow$ $l=$$\frac{{xy}}{{\sqrt {{x^2} - {y^2}} }}$  $= \frac{{AB \cdot AD}}{{\sqrt {A{B^2} - A{D^2}} }}$

In a right triangle $ABC$, right angled at $A$, on the leg $AC $ as diameter, a semicircle is described. The chord joining $A$ with the point of intersection $D$ of the hypotenuse and the semicircle, then the length $AC$ equals to

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