Question

5. In ▲ABC, <B = 115°and <C = 25°. Also, AP II BC and AQ is
the angle bisector of <BAC. What is the measure of <PAQ?

Please help Me
⭐

Answer 1

ANSWER:

∠PAQ = 45°

Correct Question:

In ΔABC, ∠B = 115°and ∠C = 25°. Also, AP | BC and AQ is the angle bisector of ∠BAC. What is the measure of ∠PAQ?

Given:

• ∠B = 115°
• ∠C = 25°
• AP | BC
• AQ is the angle bisector of ∠BAC

To Find:

• ∠PAQ

Solution:

➞ ∠PBA + ∠ABQ = 180° ( linear pair)

➞ ∠PBA + 115° = 180°

➞ ∠PBA = 180° - 115°

➞ ∠PBA = 65° ...(1)

In ΔAPB

➞ ∠P + ∠PBA + ∠PAB = 180° ( Sum of all angles of a triangle is 180°)

➞ 90° + 65° + ∠PAB = 180° { Using (1) }

➞ 155° + ∠PAB = 180°

➞ ∠PAB = 180° - 155°

➞ ∠PAB = 25° ..(2)

AQ is angle bisector of ∠BAC, so,

➞ ∠BAQ = ∠QAC

➞ ∠BAQ = x

➞ ∠QAC = x

In ΔPAC

➞ ∠APC + ∠PCA + ∠CAP = 180° ( Sum of all angles of a triangle is 180°)

[ ∠PAB + ∠BAQ + ∠QAC = CAP]

➞ ∠APC + ∠PCA + ∠PAB + ∠BAQ + ∠QAC = 180°

➞ 90° + 25° + 25° + x + x = 180°

➞ 90° + 50° + 2x = 180°

➞ 140° + 2x = 180°

➞ 2x = 180 - 140°

➞ 2x = 40°

➞ x = 40/2

➞ x = 20°

∠BAQ = 20°

∠QAC = 20° ..(3)

[ ∠PAQ = ∠BAQ + ∠PAB ]

➞ ∠PAQ = 20° + 25° {using (2) and (3)}

➞ ∠PAQ = 45°