PREVIOUSLY ASKED IN:
CTET 2026
Answer
\( 270^{\circ} \)
Explanation
In ΔPOQ, the bisected angles are \( \angle OPQ = \frac{30^{\circ}}{2} = 15^{\circ} \) and \( \angle OQP = \frac{50^{\circ}}{2} = 25^{\circ} \). The sum of angles in a triangle is 180°, so \( \angle POQ = 180^{\circ} - (15^{\circ} + 25^{\circ}) = 180^{\circ} - 40^{\circ} = 140^{\circ} \). The required value is \( 2\angle POQ - 10^{\circ} \). Substituting the value gives: \( 2(140^{\circ}) - 10^{\circ} = 280^{\circ} - 10^{\circ} = 270^{\circ} \).
Key Points
- > The sum of interior angles of any triangle is exactly 180°.
- > An angle bisector divides an angle into two equal halves.
- > The point where the angle bisectors of a triangle meet is called the Incenter.
- > There is a direct formula for the angle at the incenter: \( \angle POQ = 90^{\circ} + \frac{\angle R}{2} \).
- > Verifying with the formula: \( \angle R = 180^{\circ} - 80^{\circ} = 100^{\circ} \). So, \( \angle POQ = 90^{\circ} + 50^{\circ} = 140^{\circ} \).
- > Both methods yield exactly 140° for \( \angle POQ \).
Additional Information
Centers of a Triangle
| Center Name | Formed By Intersection of | Special Property |
|---|---|---|
| Incenter | Angle Bisectors | Equidistant from all 3 sides |
| Circumcenter | Perpendicular Bisectors | Equidistant from all 3 vertices |
| Centroid | Medians | Divides each median in a 2:1 ratio |
| Orthocenter | Altitudes (Heights) | Lies outside in an obtuse triangle |
Memory Tips
- Incenter Angle Trick: Angle formed by interior bisectors = \( 90^{\circ} + \frac{\text{Third Angle}}{2} \).
- Excenter Angle Trick: Angle formed by exterior bisectors = \( 90^{\circ} - \frac{\text{Third Angle}}{2} \).