Question

If RTQU is a straight line then ,
what is the value of b-a​

Answer 1

Answer:

Here b=134°-49°

=85°

Also ST||PR

so, a=49°

b-a=85°-49°

=36°

Hope it will help!

Answer 2

Answer:

36°

Step-by-step explanation:

Let the point where PQ intersect with ST be "F"

Now,

On straight line RTQU

∠FQU+ ∠FQT= 180°

134° + ∠FQT = 180°

∠ FQT=46°

As TS and PR are parallel lines, so they make a parellogram FTPR, and therefore ∠P = ∠R = 49°

Line PQ and TS intersect each other, so

∠RPF = ∠PFS = b = 49° ( alternate interior angles)

Now.. ∠PFS= ∠QFT = 49° (v.o. Angles are equal)

NOW, in triangleFQT

∠FQT+ ∠FTQ + ∠QFT = 180°

46° + a + 49°= 180°

a + 95°= 180°

a = 180°- 95°

a= 85° to

Now, A.C.Q.(according to question)

Value of b- a

b = 49°

a = 85°

b-a = 49°- 85°

b-a = 36° ans.

Welll, hope it helps you, it took me a lot of time to type this!

Pls! Mark me the brainliestt❤️