Find the area of the triangle $AEZ$ in the figure

Question

For reference: The angle measure $\angle ERZ=75^o$ and $EH=6$. Calculate the area of ​​the triangular region $ZEA$.(Answer:S=4)

My progress:

$OA =R\\ \triangle OAZ(equilateral)\implies S\triangle OAZ = \frac{R^2\sqrt3}{4}\\ S_{OAEZ} = \frac{\pi R^2}{6}$

S(segment AEZ) = $\frac{\pi R^2}{6} - \frac{R^2\sqrt3}{4}$

????

Answer 1

Let consider what we see in the drawing. Center of arc AHR is located in edge AR, then AR is diameter. Then triangle AHR is rectangular and EH is altitude of rectangular triangle dropped on hypotenuse. Then $$EH^2=AE\cdot ER$$

Angle AEZ is inscribed angle with corresponding arc A(not E)Z, which angular measure is $360^\circ-60^\circ=300^\circ$. Then AEZ$=150^\circ$.

In triangle REZ angle REZ$=30^\circ$, angle ERZ$=75^\circ$, then angle $EZR=75^\circ$. Then REZ is isosceles triangle with ER=EZ.

Area of triangle AEZ is $$S=\frac12 AE \cdot EZ \cdot \sin(150^\circ)=\frac14 AE\cdot EZ=\frac14 AE\cdot ER=\frac14 EH^2=9$$