Question

PQR is a triangle. The bisectors of the internal angle ∠Q and external angle ∠R intersect at M. If ∠QMR = 40°, then ∠P is:

• A. 75°
• B. 60°
• C. 65°
• D. 80°

Answer 1

The Correct Option is D

To find ∠P, we need to understand the properties of angle bisectors and the relationships between internal and external angles in a triangle.

Given: Two angle bisectors, one of the ∠Q (internal angle) and another of the ∠R (external angle), meet at point M, creating ∠QMR = 40°.

We know that the angle formed by the internal bisector and external bisector is half the difference of the internal angle and external angle:

\[ \angle QMR = \frac{1}{2}(\angle Q - \angle R) \]

Rearrange to find the difference between ∠Q and ∠R:

\[ \angle Q - \angle R = 2 \times 40^\circ = 80^\circ \]

In triangle PQR, the sum of internal angles = 180°. Use the relationships:

\( \angle R_{\text{internal}} + \angle R_{\text{external}} = 180^\circ \)

Therefore,\( \angle Q + \angle R = 180^\circ \)

Substitute the calculated difference \( \angle Q - \angle R = 80^\circ \) into the system of equations:

• \( \angle Q + \angle R = 180^\circ - \angle P \)
• \( \angle Q - \angle R = 80^\circ \)

Add the two expressions:

\( (\angle Q + \angle R) + (\angle Q - \angle R) = 180^\circ - \angle P + 80^\circ \)

\( 2\angle Q = 260^\circ - \angle P \)

\( \angle Q = 130^\circ - \frac{\angle P}{2} \)

Now substitute in the equation:\( \angle Q + 180^\circ - \angle Q = 180^\circ - \angle P \)

\( \angle R = 130^\circ - \frac{\angle P}{2} \)

\( \angle P + \angle Q + \angle R = 180^\circ \)

Plugging in the expression for \(\angle Q\) and \(\angle R\),\

\( \angle P + (130^\circ - \frac{\angle P}{2}) + (50^\circ + \frac{\angle P}{2}) = 180^\circ \)

Solving the above gives us:

\( \angle P = 80^\circ \)

Thus, ∠P is 80°.