Question

slove question no. 6 ​

Answer 1

PQ is parallel to RS.

angle PAB = angel ABC ........ alternate angles

so angle ABC = 70°.

angle ACB = 180° - 100° ....... linear pair

so, angle ACB = 80°.

Angle ABC + Angle ACB + Angle BAC = 180° ........[ SUM OF ALL ANGLES OF A TRIANGLE IS 180°

70° + 80° + Angle BAC = 180°

150° + Angle BAC = 180°

Angle BAC = 180° - 150°

= 30°

Angle CAQ = Angle ACB...... ALTERNATE ANGLES

SO angle CAQ = 80°..

Answer 2

Given,

PQ||RS

∠PAB = 70°

∠ACS = 100°

Since,

PQ || RS

And, AB is a transversal

∴∠PAB = ∠ABC = 70° (alterante angles)

Now,

∠CAQ + ∠ACS = 180° (Co-interior angles)

∠CAQ + 100° = 180°

∠CAQ = 180° - 100°

∠CAQ = 80°

Since, PQ is a line,

∴∠PAB + ∠BAC + ∠CAQ = 180°(Linear pair)

70° + ∠BAC + 80° = 180°

150° + ∠BAC = 180°

∴∠BAC = 180° - 150°

∠BAC = 30°

So,finally,

∠ABC = 110°

∠BAC = 30°

∠CAQ = 80°...