Triangle

**totalnewbie** · March 21st 2006, 07:26 AM

What is the value of angle A and C in degrees ?

**earboth** · March 21st 2006, 10:13 AM

Originally Posted by totalnewbie

What is the value of angle A and C in degrees ?

Hello,

because AB = BC then AB = BD and BC = BD because the Points A, C, D are points of the semicircle of Thales. B is the centre of this circle.

Triangle ADB is an isosceles triangle, therefore the angle at A has a value of 30°.

The angle at D is a right one. Therefore the angle at C has a value of 60°.

Hope that this helps a little bit.

Greetings

EB

**ThePerfectHacker** · March 21st 2006, 01:33 PM

I like the problem

I solved it with some trigonometry.
By law of sines we have,
$\frac{\sin A}{BD}=\frac{\sin 30^o}{AB}$ (1)
$\frac{\sin C}{BD}=\frac{\sin 60^o}{BC}$ (2)
Invert (2) to get,
$\frac{BD}{\sin C}=\frac{BC}{\sin 60^o}$
Multiply (1) and (2) to get, remember $AB=BC$
Thus,
$\frac{\sin A}{\sin C}=\frac{\sin 30^o}{\sin 60^o}$
Thus,
$\frac{\sin A}{\sin C}=\frac{\sqrt{3}}{3}$
But,
$\sin C=\cos (90^o-C)=\cos A$
Thus,
$\frac{\sin A}{\cos A}=\frac{\sqrt{3}}{3}$
Thus,
$\tan A=\frac{\sqrt{3}}{3}$
Thus,
$A=60^o$
Thus,
$C=30^o$

$\mathbb{Q}.\mathbb{E}.\mathbb{D}$

**c_323_h** · March 22nd 2006, 08:33 AM

would this work? i extended some sides and introduced parallel lines to derive the angles. not a very practical way. excuse my sloppy handwriting.

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