Question #02b52

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Question #02b52

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Answer 1 · 2017-03-17T04:12:55

see explanation.

Explanation:

Case 1.
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$sin ∠ C A D = sin A = \frac{C D}{A C}$ $sinangleCAD=SinA=(CD)/(AC)$
$\Rightarrow C D = A C \cdot sin A = b \cdot sin A$ $=> CD=AC*sinA=b*sinA$ --- (1)

$sin ∠ C B D = sin B = \frac{C D}{B C}$ $sinangleCBD=SinB=(CD)/(BC)$
$\Rightarrow C D = B C \cdot sin B = a \cdot sin B$ $=> CD=BC*SinB=a*sinB$ ---(2)

(1) = (2), $\Rightarrow b \cdot sin A = a \cdot sin B$ $=> b*SinA=a*SinB$

$\Rightarrow \frac{sin A}{a} = \frac{sin B}{b}$ $=> SinA/a=sinB/b$

Case 2.
enter image source here

$sin ∠ C A D = sin A = \frac{C D}{A C}$ $sinangleCAD=SinA=(CD)/(AC)$
$\Rightarrow C D = A C \cdot sin A = b \cdot sin A$ $=> CD=AC*sinA=b*sinA$ --- (1)

$∠ C B A = ∠ B$ $angleCBA=angleB$
$∠ C B D = 180 - ∠ B$ $angleCBD=180-angleB$
$sin ∠ C B D = sin (180 - B) = sin B = \frac{C D}{B C}$ $sinangleCBD=Sin(180-B)=sinB=(CD)/(BC)$ ,
$\Rightarrow C D = B C \cdot sin B = a \cdot sin B$ $=> CD=BC*SinB=a*sinB$ ------(2)

(1) = (2), $\Rightarrow b \cdot sin A = a \cdot sin B$ $=> b*SinA=a*SinB$

$\Rightarrow \frac{sin A}{a} = \frac{sin B}{b}$ $=> SinA/a=sinB/b$

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Question #02b52

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