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

1. Diagram and Definitions:

• Draw triangle PQR.
• Draw the internal angle bisector of ∠Q, and let it intersect the external angle bisector of ∠R at point M.
• We are given that ∠QMR = 40°.

2. Angle Relationships:

• Let ∠Q = 2x. Then, ∠MQr = x (where Qr is a line segment of Q).
• Let the external angle at R be ∠R_. Let ∠R_ext = 2y. Then, ∠MRr = y (where Rr is a line segment of R).
• In triangle QMR, we have ∠QMR + ∠MQr + ∠MRr = 180°.
• Substituting the given values, 40° + x + y = 180°.
• This gives us x + y = 140°.

3. External Angle Property:

• The external angle at R (∠R_ext) is equal to the sum of the two opposite internal angles of triangle PQR, i.e., ∠R_ext = ∠P + ∠Q.
• We have 2y = ∠P + 2x.

4. Relating Angles:

• From x + y = 140°, we get y = 140° - x.
• Substituting this into 2y = ∠P + 2x, we get 2(140° - x) = ∠P + 2x.
• This simplifies to 280° - 2x = ∠P + 2x.
• Rearranging, we get ∠P = 280° - 4x.

5. Applying Triangle Properties:

In triangle PQR, we have ∠P + ∠Q + ∠R = 180°.

∠R = 180° - ∠R_ext = 180° - 2y.

Substituting ∠Q = 2x and ∠R = 180° - 2y into ∠P + ∠Q + ∠R = 180°, we get:

• ∠P + 2x + 180° - 2y = 180°
• ∠P + 2x - 2y = 0
• ∠P = 2y - 2x
• ∠P = 2(y-x)

We also know y = 140 - x

∠P = 2(140-x-x) = 2(140-2x) = 280 - 4x

From x+y=140, we also have y-x=140-2x

Since ∠QMR = 40, we know ∠P = 2*∠QMR

∠P = 2 * 40° = 80°.

Therefore, ∠P = 80°.

The correct answer is Option 4.