# TrigonometryFinding the Area of Shaded Part

#### karush

##### Well-known member
find the area of the shaded part

first I tried to find out the area of $OAPB$ which can be found by adding triangles $OAP$ and $OBP$ which are similar, $AP$ and $BP$ are perpendicular to the radius and tangent to the circle

$BP = 12\tan{37.5^0}=9.2 cm^2$

area $\Delta OBP= \frac{1}{2}(12)(9.2)=52.2 cm^2$

area $OAPD = 2(52.2)= 110.4 cm^2$

area of sector $\frac{75}{360}\pi 12^2 = 94.2 cm^2$

so shaded area $= 110.4 - 92.2 = 16.2 cm^2$

just seeing if this is OK i went over it quite a few times...
no ans given ....

#### MarkFL

Area of $\displaystyle OAPB$ is:
$\displaystyle 15\cdot12\sin(37.5^{\circ})$
Subtract the area of sector $\displaystyle OAB$:
$\displaystyle A_S=15\cdot12\sin(37.5^{\circ})-\frac{1}{2}12^2\cdot\frac{5\pi}{12}=180\sin(37.5^{\circ})-30\pi\approx15.329277613875924$