Question

PAQ is a tangent to the circle with center O at a point A as shown in figure. If $\angle OBA={35}^{\circ }$, find the value of $\angle BAQ$ and $\angle ACB$. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

OB = OA (radius)

$\Rightarrow \angle OBA=\angle OAB={35}^{\circ }$

From $△OAB,\angle OBA+\angle OAB+\angle AOB={180}^{\circ }$

${35}^{\circ }+{35}^{\circ }+\angle AOB={180}^{\circ }$

$\angle AOB={110}^{\circ }$

But $\angle ACB=\frac{1}{2}\angle AOB$

$\Rightarrow \angle ACB=\frac{1}{2}\left({110}^{\circ }\right)$

$\angle ACB={55}^{\circ }$

As the tangent at any point of a circle is perpendicular to the radius through the point of contact,

$\angle OAQ={90}^{\circ }$

$\angle BAQ=\angle OAQ-\angle OAB$

$\angle BAQ={90}^{\circ }-{35}^{\circ }$

$\angle BAQ={55}^{\circ }$

Therefore $\angle ACB={55}^{\circ }$ and $\angle BAQ={55}^{\circ }$