# Triangle

• Mar 21st 2006, 07:26 AM
totalnewbie
Triangle
What is the value of angle A and C in degrees ?
• Mar 21st 2006, 10:13 AM
earboth
Quote:

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
• Mar 21st 2006, 01:33 PM
ThePerfectHacker
I like the problem :D
I solved it with some trigonometry.
By law of sines we have,
$\displaystyle \frac{\sin A}{BD}=\frac{\sin 30^o}{AB}$ (1)
$\displaystyle \frac{\sin C}{BD}=\frac{\sin 60^o}{BC}$ (2)
Invert (2) to get,
$\displaystyle \frac{BD}{\sin C}=\frac{BC}{\sin 60^o}$
Multiply (1) and (2) to get, remember $\displaystyle AB=BC$
Thus,
$\displaystyle \frac{\sin A}{\sin C}=\frac{\sin 30^o}{\sin 60^o}$
Thus,
$\displaystyle \frac{\sin A}{\sin C}=\frac{\sqrt{3}}{3}$
But,
$\displaystyle \sin C=\cos (90^o-C)=\cos A$
Thus,
$\displaystyle \frac{\sin A}{\cos A}=\frac{\sqrt{3}}{3}$
Thus,
$\displaystyle \tan A=\frac{\sqrt{3}}{3}$
Thus,
$\displaystyle A=60^o$
Thus,
$\displaystyle C=30^o$

$\displaystyle \mathbb{Q}.\mathbb{E}.\mathbb{D}$
• Mar 22nd 2006, 08:33 AM
c_323_h
would this work? i extended some sides and introduced parallel lines to derive the angles. not a very practical way. excuse my sloppy handwriting.