Question

In a triangle angle A is x plus 10, angle B is 3x plus 5 ,angle C is 2 x plus 15​

Answer 1

Step-by-step explanation:

In a triangle ∆ABC,

angle A = x + 10

angle B = 3x + 5

angle C = 2x + 15

angle A + angle B + angle C = 180°

x + 10 + 3x + 5 + 2x + 15 = 180°

6x + 40 = 180°

6x = 180° - 40

6x = 140

Answer 2

Given :-

In a triangle ΔABC -

∠A = ( x + 10 )°

∠B = ( 3x + 5 )°

∠C = ( x + 15 )°

To Find :-

Angles of the triangle.

Solution :-

❍ To calculate the angles of the triangle we must know the angle sum property of triangles ::

Sum of all angles of a triangle = 180°

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Finding the value of x :-

• By substituting the values in the angle sum property of triangles.

⤳ ∠A + ∠B + ∠C = 180°

⤳ ( x + 10 )° + ( 3x + 5 )° + ( 2x + 15 )° = 180°

⤳ x + 10° + 3x + 5° + 2x + 15° = 180°

⤳ 6x + 30° = 180°

⤳ 6x = 180° - 30°

⤳ 6x = 150°

⤳ x = 150°/6

⤳ x = 25°

Finding the angles :-

• By substituting the value of x in the values given to us.

⤳ ∠A = ( x + 10 )°  

⤳ ∠A = 25° + 10°

⤳ ∠A = 35°

⤳ ∠B = ( 3x + 5 )°

⤳ ∠B = ( 25° × 3 ) + 5°

⤳ ∠B = 75° + 5°

⤳ ∠B = 80°

⤳ ∠C = ( 2x + 15 )°

⤳ ∠C = ( 2 × 25° ) + 15°

⤳ ∠C = 50° + 15°

⤳ ∠C = 65°

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•  Henceforth, the angles of the triangle are 35°, 80° and 65°