In an isosceles triangle, $A B = A C, ∠ A = (2 x + 40)^{o}$ and $∠ B = (4 x + 30)^{o}$ . Find $∠ A, ∠ B$ , and $∠ C$

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Solution

Given $A B = A C$
$∠ A = {(2 x + 40)}^{\circ} ∠ B = {(4 x + 30)}^{\circ}$
we have $∠ B = ∠ C (∵ A B = A C)$
and sum of all the angles of a triangle $= 180^{\circ}$
$∴ ∠ A + ∠ B + ∠ C = 180^{\circ} {(2 x + 40)}^{\circ} + {(4 x + 30)}^{\circ} + {(4 x + 30)}^{\circ} = 180^{\circ} \Rightarrow 10 x + 100^{\circ} = 180^{\circ} \Rightarrow 10 x = 80^{\circ} \Rightarrow x = 8^{\circ} ∴ ∠ A = {(2 x + 40)}^{\circ} = {(2 \times 8 + 40)}^{\circ} = 56^{\circ} ∠ B = ∠ C = {(4 x + 30)}^{\circ} = {(4 \times 8 + 30)}^{\circ} = 62^{\circ} ∠ A = 56^{\circ} ∠ B = 62^{\circ} ∠ C = 62^{\circ}$

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In an isosceles triangle, AB=AC,∠A=(2x+40)o and ∠B=(4x+30)o. Find ∠A,∠B, and ∠C

In an isosceles triangle, $A B = A C, ∠ A = (2 x + 40)^{o}$ and $∠ B = (4 x + 30)^{o}$ . Find $∠ A, ∠ B$ , and $∠ C$