Question

In the figure above, triangle $ABC$ is inscribed in the circle with center $O$ and diameter $\overline{AC}$. If $AB=AO$, what is the degree measure of $\angle ABO$?

• A. ${15}^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${60}^o$
• E. ${90}^o$

Answer 1

Answer: (D)

${60}^o$

From the figure,

As $△ABC$ is inscribed in the circle, the points ${}^{\prime }{A}^{\prime }$, ${}^{\prime }{B}^{\prime }$, ${}^{\prime }{C}^{\prime }$ lie on the circumference of the circle

As ${}^{\prime }{O}^{\prime }$ is the center of the circle. $OA$, $OB$, $OC$ are equal to the radius of the circle.

Hence, $OA$ $=$ $OB$ $=$ $OC$

In $△ABO$,

As $AB$ $=$ $AO$ ..... (given)

From above,

In $△ABO$, $OA$ $=$ $AB$ $=$ $OB$

As three sides of a triangle are equal, $△ABO$ is equilateral.

In an equilateral triangle, each angle is equal to ${60}^0$.

Hence, $\angle ABO$ $=$ $\angle BOA$ $=$ $\angle OAB$ $=$ ${60}^0$.

Therefore, degree measure of $\angle ABO$ is ${}^{\prime }$${60}^0$${}^{\prime }$.