Question

In the figure given below, ABOP is a rectangle and O is the centre of the circle. It is also given that AB = BC and the measure of the $\angle$ABC is 60${}^{\circ }$. Find the measure of the angle OPN.

• A. 30°
• B. 15°
• C. 20°
• D. 25°

Answer 1

The correct option is B 15°

AB = BC and $\angle$ABC=60${}^{\circ }$. Therefore, ΔABC is an equilateral triangle

Now see that ABOP is a rectangle.

And $\angle$BAN=60${}^{\circ }$, Therefore, $\angle$NAP=90${}^{\circ }$ − 60${}^{\circ }$ = 30${}^{\circ }$

And $\angle$ANP = $\frac{1}{2}$ * 90 = 45${}^{\circ }$

Now in ΔANP,

$\angle$ NPA=180${}^{\circ }$−45${}^{\circ }$−30${}^{\circ }$=105${}^{\circ }$

And hence $\angle$ NPO = $\angle$ NPA - $\angle$ OPA = 105${}^{\circ }$- 90${}^{\circ }$ = 15${}^{\circ }$

Shortcut:
ABC is an equilateral triangle and ABM is a 30 – 60 -90 triangle (M being the point of intersection of AN and the circle). OMN is also 30${}^{\circ }$. MOP = 90${}^{\circ }$, and MNP = 45${}^{\circ }$; MPO = PMO = 45${}^{\circ }$. NPO = 180${}^{\circ }$ – 75${}^{\circ }$ – 45${}^{\circ }$ – 45${}^{\circ }$ = 15${}^{\circ }$