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#1

$Old$ November 5th, 2009, 10:26 AM

ukorov $ukorov is offline$

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$Default$ problem with triangular geometry

referring to the attached figure:
In $\triangle ABC$ ,
D is mid-point of BC,
$\angle ABD = \angle DAC = x$
angle ADB = $45^o$

Need desperate help to find the value of x. Thanks.

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$problem-triangular-geometry-ques-10.jpg$

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$Old$ November 5th, 2009, 10:47 AM

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The triangles ABC, DAC are similar (equal angles). Deduce that $AC = \sqrt2DC$ . Then apply the sine rule in triangle ADC to find that $\sin x = 1/2$ .

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$Old$ November 5th, 2009, 11:06 AM

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Originally Posted by Opalg $View Post$

The triangles ABC, DAC are similar (equal angles). Deduce that $AC = \sqrt2DC$ . Then apply the sine rule in triangle ADC to find that $\sin x = 1/2$ .

alright that sounds easy, so x = 30 degrees

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