Question

in the isosceles triangle abc angle a and angle b are equal . angle acd in an exterior angle of triangle abc the measures of angle acb and angle acd are (3x-17°) and (8x+10) respectively find the measures of angle acb and angle acd also find the measures of angle a and angle b​

step by step answers

Answer 1

Answer:

∠ acb = 26

∠ acb = 98

∠ a = 77

∠ b = 77

Step-by-step explanation:

∠ acb + ∠ acd = acd (straight line)

(3x-17°) + (8x+10°) = 180°

3x + 8x = 180 + 17 - 10

11x = 187

x = 187/11

x = 17

∠ acb = (3x-17°)

                = 3(11) - 17

                = 33 -17

∠ acb = 26

∠ acd = (8x+10°)

                 = 8(11) + 10

                 = 88 + 10

∠ acb = 98

∠ a + ∠ b + ∠ acb = Δ abc

2∠ a + ∠ acb = Δ abc      (∠ a = ∠ b)

2∠ a + 26 = 180

2∠ a = 180 - 26

2∠ a = 154

∠ a = 154/2

∠ a = 77

as ∠ a = ∠ b

∠ b = 77