• Jan 7th 2007, 07:30 AM
r_maths
http://i1.tinypic.com/4gxilvm.jpg

The diagram shows a kite whose leading diagonal forms the diameter of a circle. Find the exact value of sin XOY

What I want to know is, do I need to find the diameter, if so, how?
• Jan 7th 2007, 08:07 AM
ThePerfectHacker
Quote:

Originally Posted by r_maths
http://i1.tinypic.com/4gxilvm.jpg

The diagram shows a kite whose leading diagonal forms the diameter of a circle. Find the exact value of sin XOY

What I want to know is, do I need to find the diameter, if so, how?

That means the inscribed angle X0Y is 90 degree.
Thus, sin 90 = 1
• Jan 7th 2007, 08:30 AM
r_maths
Quote:

Originally Posted by ThePerfectHacker
That means the inscribed angle X0Y is 90 degree.
Thus, sin 90 = 1

my bad, the sides are not equal so you cant 360/4.
root 5 & 3 are the length of the sides of the kite

the answer is 3 root5 / 7
i dont know how to arrive at that.
(just to let you know, no calculator allowed)
• Jan 7th 2007, 08:34 AM
Soroban
Hello, r_maths!

Quote:

The diagram shows a kite whose leading diagonal forms the diameter of a circle.
Find the exact value of $\displaystyle \sin(XOY)$.

Look at half of the kite . . .
Code:

      Z       *       | *  _       |  *√5       |    *   __ |  90° * Y   √14 |      *       |    *       |    *       |  * 3       |θ *       | *       |*       *       O

Since $\displaystyle ZO$ is a diameter, $\displaystyle \angle Y = 90^o.$

Using Pythagorus, we find that: .$\displaystyle ZO = \sqrt{14}$

Let $\displaystyle \theta = \angle ZOY$
Then: .$\displaystyle \sin\theta = \frac{\sqrt{5}}{\sqrt{14}},\;\cos\theta = \frac{3}{\sqrt{14}}$

Then: .$\displaystyle \sin(XOY) \:=\:\sin(2\theta) \:=\:2\sin\theta\cos\theta \:=\:2\left(\frac{\sqrt{5}}{\sqrt{14}}\right)\left (\frac{3}{\sqrt{14}}\right) \:=\:\frac{3\sqrt{5}}{7}$

• Jan 7th 2007, 08:48 AM
r_maths
Quote:

Originally Posted by Soroban
Hello, r_maths!

Look at half of the kite . . .
Code:

      Z       *       | *  _       |  *√5       |    *   __ |  90° * Y   √14 |      *       |    *       |    *       |  * 3       |θ *       | *       |*       *       O

Since $\displaystyle ZO$ is a diameter, $\displaystyle \angle Y = 90^o.$

how did you work out that angle was 90 ?