Let A B C be a triangle in which A B=B C. Let X be a point on A B, such that A   X :   X   B = A   B :   A   X   . If A   C = A   X , then the measure of ∠ A B C equals

Step-by-step solution

lt is given that in Δ A B C , A B = B C and A X X B = A B A X = 1 x ( say ) ⇒ A X = x · A B and X B = x A X ∴ X B = x 2 ⋅ A B ∵ A B = A X + X B = x ⋅ A B + x 2 ⋅ A B ⇒ x 2 + x − 1 = 0 ⇒ x = − 1 ± 1 + 4 2 = ± 5 − 1 2 ∴ x > 0 , so x = 5 − 1 2 Now, cos θ = A B 2 + B C 2 − A C 2 2 ( A B ) ( B C ) ⇒ cos θ = 2 A B 2 − A X 2 2 A B 2 [ ∵ A B = B C ] = 2 A B 2 − x 2 ⋅ A B 2 2 A B 2 = 2 − x 2 2 = 2 − 5 − 1 2 2 2 = 8 − ( 5 + 1 − 2 5 ) 8 = 2 5 + 2 8 = 5 + 1 4 = cos 36 ∘ So , θ = ∠ A B C = 36 ∘