A line segment AB = 6 cm. And a semicircle with AB as the diameter

Question

A line segment $AB = 6 cm.$ And a semicircle with $AB$ as the diameter. And point $P$ at arc $AB$, $\angle ABP = x$. What is area of $\triangle ABP$ in terms of $x$ ?

My attempt : Area = $3PB\sin x$ $AB^2 + PB^2 = AP^2 + 2AB.PB\cos x$

Answer 1

$\triangle ABP$ is a right triangle with hypotenuse $6$ (angle within a semicircle is a right angle).

Length of side adjacent to $\angle ABP = 6\cos x$.

Area of triangle = $\frac 12 mn\sin\theta$, where $m$ and $n$ are two sides and $\theta$ is the enclosed angle.

So area of triangle = $\frac 12 (6)(6\cos x)\sin x = 18\sin x\cos x = 9 \sin 2x$.