I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
Here is the problem:
The diagram shows a circle of radius 8 metres. The points ABCD lie on the circumference of the circle.
BC = 14 m, CD = 11.5 m, AD = 8 m, $\angle ADC$ = 104°, and $\angle BCD$ = 73°.
a. Find AC.
AC is the diameter, right? So you would have double the radius which is 8 metres. So AC = 16?
b. (i) Find $\angle ACD$
(ii) Hence, find $\angle ACB$
If angle C is 73°, would ACD be half of that?
And then ACB is the other half?
c. Find the area of triangle ADC.
I believe the area of the triangle is 1/2 x base x height. So, 1/2 x 8 x 11.5?
d. Hence or otherwise, find the total area of the shaded regions.
This is confusing and I do not know how to do this part.
Best Answer
Since $\angle ACD$ subtend $104^\circ$, AC is NOT a diameter.
Use Cosine rule to find AC $$AC^2=AD^2+DC^2-2AD\cdot DC\cdot\cos104^\circ$$ $$AC^2=8^2+11.5^2-2\times8\times11.5\times(-0.241)$$ $$ AC=15.51m$$
Then use Sine rule to find $\angle ACD$ $$\frac{\sin{\angle ACD}}{8}=\frac{\sin{\angle ADC}}{AC}$$ $$\frac{\sin{\angle ACD}}{8}=\frac{\sin{\angle 104}}{15.51}$$ $$sin\angle ACD=0.5$$ $$\angle ACD=30^\circ$$ So $\angle ACB=73^\circ-30^\circ=43^\circ$
Best way to find $\triangle ADC $ area is to choose $AC$ as the base then find height using $DC\cdot\sin\angle ACD$ (i.e perpendicular distance from D to side AC)
Area of $\triangle ADC=\frac12\times AC\times11.5\times \sin30 =44.59m^2$
Similarly, find Area of $\triangle ABC$,
$$\triangle ABC= \frac12\times AC\times BC\times \sin \angle ACB$$ $$= \frac12\times 15.51\times 14\times \sin43^\circ$$ $$= 74.18m^2$$
Shaded Area=$ \pi\times8^2-(\triangle ABC +\triangle ADC)$