In the given figure- POR is a line- The value of a is a 40-b 45 -c 55- d 60-

Since POR is a line. So the sum of angle forming linear pair is 180^∘∴∠ POQ + ∠ ROQ = 180^∘ ⇒ (3a+5)^∘ + (2a 25^∘) = 180^∘ ⇒ 3a+5^∘+2a 25^∘ = 180^∘ ⇒ 5a 20^∘ = ...

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In the given ...

Question

In the given figure, POR is a line. The value of a is

(a) $40^{\circ}$
(b) $45^{\circ}$
(c) $55^{\circ}$
(d) $60^{\circ}$

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Solution

Since POR is a line. So the sum of angle forming linear pair is $180^{\circ} ∴ ∠ P O Q + ∠ R O Q = 180^{\circ} \Rightarrow (3 a + 5)^{\circ} + (2 a - 25^{\circ}) = 180^{\circ} \Rightarrow 3 a + 5^{\circ} + 2 a - 25^{\circ} = 180^{\circ} \Rightarrow 5 a - 20^{\circ} = 180^{\circ} \Rightarrow 5 a = 200^{\circ} \Rightarrow a = \frac{200^{\circ}}{5} \Rightarrow a = 40^{\circ}$

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Similar questions

Q. In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD = ?
(a) 45°
(b) 55°
(c) 60°
(d) 75°

Q. In the given figure, PQ and PR are tangents drawn from P to a circle with centre O. If ∠OPQ = 35°, then

(a) a= 30°, b= 60°

(b) a= 35°, b = 55°

(c) a= 40°, b = 50°

(d) a= 45°, b = 45°

Q. In the given figure, what is the value of x?

(a) 35

(b) 45

(c) 50

(d) 60

Question 13
In the given figure, POR is a line. The value of a is

(a) $40^{\circ}$
(b) $45^{\circ}$
(c) $55^{\circ}$
(d) $60^{\circ}$

In given figure, ABCD is a rhombus. If ∠ O A B = 40^{\circ}, then find the value of x.

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In the given figure, POR is a line. The value of a is (a) 40∘ (b) 45∘ (c) 55∘ (d) 60∘

Since POR is a line. So the sum of angle forming linear pair is 180∘∴∠POQ+∠ROQ=180∘⇒(3a+5)∘+(2a−25∘)=180∘⇒3a+5∘+2a−25∘=180∘⇒5a−20∘=180∘⇒5a=200∘⇒a=200∘5⇒a=40∘

In the given figure, POR is a line. The value of a is

(a) $40^{\circ}$
(b) $45^{\circ}$
(c) $55^{\circ}$
(d) $60^{\circ}$