Question

one angle of a triangle is 61 degrees. the other two are in the ratio one whole 1/2 : one whole 1/3​

Answer 1

Step-by-step explanation:

In ∆ ABC,

Let ∠A = 61°

But ∠A + ∠B + ∠C = 180°

(Angles of a triangle)

⇒ 61° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° - 61° = 119°

But ∠B : ∠C = 1.1/2 : 1.1/3 = 3/2 : 4/3

= (9 : 8)/6 = 9 : 8

Let ∠B = 9x and ∠C = 8x,

then, 9x + 8x = 119°

⇒ 17x = 119°

⇒ x = 119°/17 = 7°

∠B = 9x = 9 × 7° = 63°

C = 8x = 8 x 7° = 56°

Answer 2

In ∆ ABC,

Let ∠A = 61°

But ∠A + ∠B + ∠C = 180°

(Angles of a triangle)

⇒ 61° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° - 61° = 119°

But ∠B : ∠C = 1.1/2 : 1.1/3 = 3/2 : 4/3

= (9 : 8)/6 = 9 : 8

Let ∠B = 9x and ∠C = 8x,

then, 9x + 8x = 119°

⇒ 17x = 119°

⇒ x = 119°/17 = 7°

∠B = 9x = 9 × 7° = 63°

C = 8x = 8 x 7° = 56°

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