Question

In figure angle ABP = angle ACQ, angle BAC 68 degree. Find angle ABC, CBR, RBP and PBA

Answer 1

∵∠ ABP = ∠ACQ (given)

∴ 180° -  ∠ABP =  180° - ∠ACQ

⇒ ∠ABC =  ∠ACB

So, let ∠ABC = ∠ACB = x

∴ ∠BAC + ∠ABC + ∠ACB = 180° (sum of all the angles of a triangle is 180°)

⇒ 68° + x + x = 180°

⇒ 2x  =  180° - 68°

⇒ 2x  =  112°

∴    x   =  56°

∴ ∠ACB = ∠ABC = 56°

Now,

∠ABC + ∠CBR = 180° (Linear pair of angles)

∴ 56° + ∠CBR = 180°

⇒ ∠CBR = 180° -  56°

⇒ ∠CBR = 124°

Now,

∠RBP = ∠ABC (vertically opposite angles)

∴ ∠RBP = 56°

Similarly,

∠PBA = ∠CBR (vertically opposite angles)

∴ ∠PBA = 124°

Hope it will help you..............

Answer 2

Step-by-step explanation:

From the image you will get the way to solve this geometry.

Hope this helps you