Question

1. In figure segAB = seg AC= seg = segCQ
and <BAC =40° then find measure of angle PAQ .​

Answer 1

∴

$\underline{ \boxed {\bold {Question}}}$

• Find ∠PAQ

$\underline{ \boxed {\bold {Given \; in\; figure}}}$

• AB = PB
• AC = CQ
• segAB = seg AC (AB = AC)
• ∠BAC = 40°

$\underline{ \boxed {\bold {Important \: Concept \: To \: Be \: learned}}}$

• Angles opposite to equal sides of a triangle are equal.

• Sides opposite to equal angles of a triangle are equal.

• Sum of all interior angles of triangle is 180°

• Sum of linear pair of angles is 180°

$\underline{ \boxed {\bold {Answer}}}$

As from above concept,

∠ABC = ∠ACB

By Δ sum property,

= ∠ABC + ∠ACB + ∠BAC = 180°

= 2∠ABC + ∠BAC = 180°

= 2∠ABC + 40° = 180°

= 2∠ABC  = 160°

$\underline{ \boxed {\bold {\therefore \angle ABC = \angle ACB = 80^{\°}}}}$

Now by linear pair,

∠ABP + ∠ABC = 180°

∠ACQ + ∠ACB = 180°

$\underline{ \boxed {\bold {\therefore \angle ABP = \angle ACQ = 100^{\°}}}}$

→ Now in ΔAPB

∠PAB = ∠BPA

∠PAB + ∠BPA + ∠PBA = 180°

2∠PAB = 180° - 100

$\underline{ \boxed {\bold {\therefore \angle PAB = 40^{\°}}}}$

→ Now in ΔAQC

∠CAQ = ∠CQA

∠CAQ + ∠CQA + ∠ACQ = 180°

2∠CAQ = 180° - 100

$\underline{ \boxed {\bold {\therefore \angle CAQ = 40^{\°}}}}$

∴ ∠PAQ = ∠PAB + ∠BAC + ∠CAQ

∴ ∠PAQ = 40° + 40° + 40° = 120°

$\underline{ \boxed {\bold {\therefore \angle PAQ = 120^{\°}}}}$

Hope this helps u..../

【Brainly Advisor】