In Fig. 92, AOB is a straight line such that ∠AOC = (3x + 10)°, ∠COD = 50° and ∠BOD = (x − 8)°. The value of x is
(a) 32
(b) 36
(c) 42
(d) 52

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Solution

∠AOC + ∠COD + ∠BOD = 180^∘ [AOB is a straight line]
⇒ (3x + 10)^∘ + 50^∘+ (x − 8)^∘ = 180^∘
⇒ 3x + 10 + 50+ x − 8 = 180
⇒ 4x + 52 = 180
⇒ 4x = 128
⇒ x = 32
Hence, the correct answer is option (a).

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In Fig. 92, AOB is a straight line such that ∠AOC = (3x + 10)°, ∠COD = 50° and ∠BOD = (x − 8)°. The value of x is (a) 32 (b) 36 (c) 42 (d) 52

∠AOC + ∠COD + ∠BOD = 180∘ [AOB is a straight line] ⇒ (3x + 10)∘ + 50∘ + (x − 8)∘ = 180∘ ⇒ 3x + 10 + 50 + x − 8 = 180 ⇒ 4x + 52 = 180 ⇒ 4x = 128 ⇒ x = 32 Hence, the correct answer is option (a).

In Fig. 92, AOB is a straight line such that ∠AOC = (3x + 10)°, ∠COD = 50° and ∠BOD = (x − 8)°. The value of x is
(a) 32
(b) 36
(c) 42
(d) 52

∠AOC + ∠COD + ∠BOD = 180^∘ [AOB is a straight line]
⇒ (3x + 10)^∘ + 50^∘+ (x − 8)^∘ = 180^∘
⇒ 3x + 10 + 50+ x − 8 = 180
⇒ 4x + 52 = 180
⇒ 4x = 128
⇒ x = 32
Hence, the correct answer is option (a).