Question

35)
(2x + 251
60°
40°
3. In the given figure, PQR is a straight line. Find the value of x.
15
B.​

Answer 1

Answer:

MATHS

In the figure, PQR is a straight line. Find x. Then complete the following :

(i) ∠AQB=

(ii) ∠BQP=

(iii) ∠AQR=

1447790

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ANSWER

In the given fig., ∠PQA,∠AQB and ∠BQR forms a linear pair.

Therefore,

∠PQA+∠AQB+∠BQR=180°

(x+20°)+(2x+10°)+(x−10°)=180°

4x+20°=180°

⇒x=

4

180°−20°

=40°

Therefore,

∠AQB=2x+10°=2×40°+10°=90°

∠BQP=(x+20°)+(2x+10°)=3x+30°=3×40°+30°=150°

∠AQR=(2x+10°)+(x−10°)=3x=3×40°=120°

Answer 2

Answer:

ANSWER

In the given fig., ∠PQA,∠AQB and ∠BQR forms a linear pair.

Therefore,

∠PQA+∠AQB+∠BQR=180°

(x+20°)+(2x+10°)+(x−10°)=180°

4x+20°=180°

⇒x=4180°−20°=40°

Therefore,

∠AQB=2x+10°=2×40°+10°=90°

∠BQP=(x+20°)+(2x+10°)=3x+30°=3×40°+30°=150°

∠AQR=(2x+10°)+(x−10°)=3x=3×40°=120°