In triangle $ABC$, $AB = BC = 25$ and $AC = 40$. What is $\sin \angle ACB$?

Sine angle of <ACB = 38.68°

Step-by-step explanation:

Hello,

To solve this problem, we need a good representation of the sides and the angle.

See attached document for better illustration.

Assuming it's a right angled triangle,

AC = hypothenus

AB = opposite

BC = adjacent

AC = 40

BC = 25

AB = 25

From trigonometric ratios

Sinθ = opposite/ hypothenus

Sinθ = AB / AC

Sinθ = 25 / 40

Sinθ = 0.625

θ = sin⁻¹0.625

θ = 38.68°

Sine angle of <ACB = 38.68°

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