Question

In the given figure, O is the centre of the circle. PQ is the tangent to the circle at A. If $\angle$ PAB = 60${}^0$ then find $\angle$ABO.

• A. 30${}^0$
• B. 60${}^0$
• C. 90${}^0$
• D. Can't be found

Answer 1

The correct option is A 30${}^0$

Given, $\angle$PAB = 60${}^0$ and PQ is tangent to the circle.
Now, draw a line from 'O' to 'A'.
$\because$ 'A' is the point of contact of tangent to the circle
$\therefore$ $\angle$OAQ = $\angle$OAP = 90${}^0$
$\because$ $\angle$PAB = 60${}^0$
$\therefore$ $\angle$OAB = 90${}^0$ - 60${}^0$
= 30${}^0$
Now, in $\Delta$OAB, OB = OA [OA = OB = radius]
$\therefore$ $\angle$OAB = $\angle$OBA ($\because$ If two sides of a triangle are equal then their corresponding angles are also be equal)
$\Rightarrow$ $\angle$OBA = 30${}^0$