1. ## Kite in circle

I'm not sure how to start the question. It will need a double angle formula later.

2. hint ...

an angle inscribed in a semicircle is a right angle.

3. Hello, Stuck Man!

Yes, both double-angle identities are useful.

12. The diagram shows a kite who leading diagonal is a diameter of the circle.

Find the exact values of: . $(a)\;\sin(\angle XOY) \qquad\qquad (b)\;\cos(\angle XOY)$
Code:
                Z
* * *  _
*  *  |  *√5*
* *     |     * *
X o        |        o Y
\       |       /
*  \      |      /  *
*   \     |     /   *
*  3 \    |    / 3  *
\ θ | θ /
*     \  |  /     *
*     \ | /     *
*    \|/    *
* * *
O
As skeeter pointed out: . $\angle X =\angle Y =90^o$

Let: $\theta =\angle XOZ =\angle YOZ$

Then: . $\tan\theta \:=\:\frac{\sqrt{5}}{3} \:=\:\frac{opp}{adj}$

Since $opp = \sqrt{5},\;adj = 3\quad\Rightarrow\quad hyp = \sqrt{14}$

Hence: . $\sin\theta \,=\,\frac{\sqrt{5}}{\sqrt{14}},\;\;\cos\theta \:=\:\frac{3}{\sqrt{14}}$

$(a)\;\;\sin(XOY) \;\;=\;\;\sin(2\theta) \;\;=\;\;2\sin\theta\cos\theta \;\;=\;\;2\left(\frac{\sqrt{5}}{\sqrt{14}}\right)\ left(\frac{3}{\sqrt{14}}\right)$ . $=\;\;\frac{6\sqrt{5}}{14} \;\;=\;\;\frac{3\sqrt{5}}{7}$

$(b)\;\;\cos(XOY) \;\;=\;\;\cos(2\theta) \;\;=\;\;\cos^2\!\theta - \sin^2\!\theta \;\;=\;\;\left(\frac{3}{\sqrt{14}}\right)^2 - \left(\frac{\sqrt{5}}{\sqrt{14}}\right)^2$ . $=\;\;\frac{9}{14} - \frac{5}{14} \;\;=\;\;\frac{4}{14} \;\;=\;\;\frac{2}{7}$

4. Thanks. I didn't know about the right angle. I would probably have done the rest ok.