Question

AC is a diameter of circle O and m BOC=54, Find m BAO

Answer 1

Answer on Question #24177 – Math – Geometry

Question

AC is a diameter of circle centered at O and $m\angle BOC = 54$. Find $m\angle BAO$.

Solution

Given

$$m\angle BOC = 54{}^\circ$$

calculate

$$m\angle BOA = m\angle COA - m\angle COB = 180{}^\circ - 54{}^\circ = 126{}^\circ.$$

$AO$ is a radius, $BO$ is a radius too. Therefore, the triangle $\Delta ABO$ is isosceles.

That’s why

$$m\angle BAO = m\angle ABO.$$

The sum of the angles of a triangle is $180{}^\circ$, that is,

$$m\angle BAO + m\angle ABO + m\angle BOA = 180{}^\circ,$$ $$m\angle BAO + m\angle ABO + 126{}^\circ = 180{}^\circ,$$ $$m\angle BAO + m\angle ABO = 180{}^\circ - 126{}^\circ,$$ $$m\angle BAO + m\angle ABO = 54{}^\circ.$$

Thus, we get the system of equations $\left\{ \begin{array}{c} m\angle BAO + m\angle ABO = 54{}^\circ, \\ m\angle BAO = m\angle ABO. \end{array} \right.$

Consider

$$m\angle BAO + m\angle ABO = m\angle BAO + m\angle BAO = 2m\angle BAO = 54{}^\circ,$$

that is,

$$2m\angle BAO = 54{}^\circ,$$

dividing by 2 obtain

$$m\angle BAO = 27{}^\circ.$$

Answer: $m\angle BAO = 27{}^\circ$.

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