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

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- Mar 21st 2006, 07:26 AMtotalnewbieTriangle
What is the value of angle A and C in degrees ?

- Mar 21st 2006, 10:13 AMearbothQuote:

Originally Posted by**totalnewbie**

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 PMThePerfectHacker
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 AMc_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.