# Kite in circle

• Jan 31st 2010, 10:08 AM
Stuck Man
Kite in circle
I'm not sure how to start the question. It will need a double angle formula later.
• Jan 31st 2010, 11:58 AM
skeeter
hint ...

an angle inscribed in a semicircle is a right angle.
• Jan 31st 2010, 02:36 PM
Soroban
Hello, Stuck Man!

Yes, both double-angle identities are useful.

Quote:

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

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

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

• Feb 1st 2010, 07:44 AM
Stuck Man
Thanks. I didn't know about the right angle. I would probably have done the rest ok.