Question

TWO ANGLES OF A TRIANGLE ARE IN THE RATIO 4:5. IF THE SUM OF OF THESE ANGLES IS EQUAL TO THE THIRD ANGLE.FIND THE ANGLES

Answer 1

Let the Triangle be ABC
Given: ∠A= 4x  ; ∠B= 56x   ; ∠C= ∠A+∠B
                                              ∠C= 4x+5x= 9x
By Angle Sum Property
    ∠A+∠B+∠C= 180°
    4x+5x+9x= 180°
     18x= 180°
     x= 180/18= 10°
∵  ∠A= 4x= 4(10)= 40°
    ∠B= 5x= 5(10)= 50°
    ∠C= 9x= 9(10)= 90°

Answer 2

SOLUTION

Let the triangle (Δ) be ABC

Given : ∠A = 4x

∠B = 5x

∠C = ∠A+∠B

∠C= 4x + 5x = 9x

                                               

Here we use angle sum property

⇒ ∠A + ∠B + ∠C = 180°

⇒ 4x + 5x + 9x

⇒ 18x = 180°

⇒ x = $\dfrac{180^\circ}{18}$

⇒ x = 10°

                                               

Let us verify that

to verify our answer  we have to put 10°  at the place of 'x'

let's do that

                                             

⇒  ∠A + ∠B + ∠C = 180°

⇒ 4x + 5x + 9x  = 180°

⇒ 4×10 + 5×10 + 9×10 = 180°

⇒ 40 + 50 + 90 = 180°

⇒ 90 + 90 = 180°

⇒ 180°  = 180°

                                           

∠A = 4x  = 4×10 = 40°

∠B = 5x  = 5×10 = 50°

∠C = ∠A+∠B

∠C= 4x + 5x = 9x  = 9×10 = 90°

                                             

Therefore angles of triangle are 40° , 50°  and 50°