Question

The figure shows a circle with center P and inscribed isosceles Triangle ABC. If AC has the same length as the radius of the circle, what is the measure of angle ABC

Answer 1

Answer:

<ABC = $30^{0}$ (The central angle of a circle is twice any inscribed angle subtended by the same arc).

Step-by-step explanation:

From the diagram, ABC is an inscribed isosceles triangle. But the radius of the circle equals the length AC.

Join P to A and C to form an equilateral triangle. An equilateral triangle has equal sides and angles. So, the value of each interior angle of the equilateral triangle is;

Sum of angle in a triangle = $180^{0}$

So that each interior angle = $\frac{180^{0} }{3}$

                                            = $60^{0}$

The value of each interior angle of the triangle is $60^{0}$. Thus, <APC = $60^{0}$.

⇒ <ABC = $30^{0}$ (The central angle of a circle is twice any inscribed angle subtended by the same arc.)