# Trig Functions (Angles subtended?)

• May 29th 2007, 09:55 PM
MathIsPower
Trig Functions (Angles subtended?)
Done
• May 29th 2007, 10:29 PM
earboth
Quote:

Originally Posted by MathIsPower
Hey everyone I really need help with this question for my homework and I'm having a hard time with it.

Q- The angle subtended at the center by the arc PQ is 3Pi/8 radians. R is the midpoint of arc PQ. Find the shaded area.

http://olympic.txc.net.au/AK/help.JPG (Diagram)

Any help will be much appreciated please :) Thanks in advance!

Hello,

the shaded area can be calculated by:

area of sector - area of right triangle

The angle(ROP) = $\frac{3 \pi}{16}$

Thus the area of the sector is: $\frac{A_{sector}}{\pi r^2}=\frac{\frac{3 \pi}{16}}{2\pi} \ \Longleftrightarrow \ A_{sector}=\frac{3}{32} \pi r^2$

The area of the right triangle is $A_{\text{right triangle}}=\frac{1}{2} \cdot {base} \cdot {height}$

If r is the length of the radius (here r = 4 cm) then the length of the base: $b=r \cdot \cos \left(\frac{3 \pi}{16} \right)$ and

the length of the height is: $h=r \cdot \sin \left(\frac{3 \pi}{16} \right)$

Therefore the shaded area is:

$A_{\text{shaded area}}=\frac{3}{32} \pi r^2 - \frac{1}{2} \cdot r \cdot \cos \left(\frac{3 \pi}{16} \right) \cdot r \cdot \sin \left(\frac{3 \pi}{16} \right)$

Plug in the values you know to calculate the area.

(For confirmation only: I've got A = 1.017 cm²)
• May 30th 2007, 01:29 PM
MathIsPower
Thanks Heaps!