Question

In the given figure, ABCD is a rectangle, OB is the radius, PB is the tangent at point B and $\angle OBC={30}^{\circ }$. Then, the value of x is___.

• A. ${90}^{\circ }$
• B. ${30}^{\circ }$
• C. ${110}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$

Given: $\angle$ OBC = ${30}^{\circ }$
Consider, $\angle$ ABC
$\angle$ ABC = $\angle$ OBC + $\angle$ OBA
$\Rightarrow$ $\angle$ ABC = ${30}^{\circ }$ + $\angle$ OBA ---- (1)
$\because$ ABCD is a rectangle
$\therefore$ $\angle$ ABC = ${90}^{\circ }$
$\Rightarrow$ ${90}^{\circ }$ = ${30}^{\circ }$ + $\angle$ OBA [from (1)]
$\Rightarrow$ $\angle$ OBA = ${90}^{\circ }$ - ${30}^{\circ }$
$\therefore$ $\angle$ OBA = ${60}^{\circ }$

Now,
Consider $△$ OAB
$\because$ OB = OA [radius]
By Theorem - Angle opposite to equal sides of a triangle are equal.
$\therefore$ $\angle$ OAB = $\angle$ OBA = ${60}^{\circ }$ ----(2)
By Theorem- Sum of angles of a triangle = ${180}^{\circ }$
$\therefore$ $\angle$ OAB + $\angle$ OBA + $\angle$ AOB = ${180}^{\circ }$
$\Rightarrow$ 2$\angle$ OAB + $x$ = ${180}^{\circ }$ [from (2)]
$\Rightarrow$ 2$\times$${60}^{\circ }$ + $x$ = ${180}^{\circ }$
$\Rightarrow$ $x$ = ${180}^{\circ }$ - ${120}^{\circ }$
$\therefore$ $x$ = ${60}^{\circ }$