Thread: Trig functions question

1. Trig functions question

Two discs,each of length 11.6 cm are laid on a table with their centers 14cm apart.
find the area common to the two discs?
i used the formula 1/2(r^2)theta to find the area of sector and then subtract the area of triangle from it 1/2(r^2)sin.theta
the values i get are 60.14 for area of sector
and 105.62 (i'm in radians)
whats going wrong here.are my value right?

2. Hello kandyfloss
Originally Posted by kandyfloss
Two discs,each of length 11.6 cm are laid on a table with their centers 14cm apart.
find the area common to the two discs?
i used the formula 1/2(r^2)theta to find the area of sector and then subtract the area of triangle from it 1/2(r^2)sin.theta
the values i get are 60.14 for area of sector
and 105.62 (i'm in radians)
whats going wrong here.are my value right?
Have a look at the diagram I've attached.

I assume that it's the radius of the discs that is 11.6 cm. So in the $\displaystyle \triangle ABO$:
$\displaystyle \cos \theta = \frac{7}{11.6}$

$\displaystyle \Rightarrow \theta = 0.9230$ radians

and the area of the overlap is 4 times the area of the shape bounded by the straight lines AB and BP and the arc AP; which is:
$\displaystyle 4\times\big(\tfrac12\cdot11.6^2\cdot0.9230 - \tfrac12\cdot11.6\cdot7\cdot\sin(0.9230)\big)$ (that's the area of sector OAP minus the area of triangle OAB)
$\displaystyle =118.9 \text{ cm}^2$, if my arithmetic is correct

3. Yes 11.6 was the radius and the answer is right 118.9.
i tried doing the same thing but by using the angle AOC (pls see the picture)
as 90 deg..because it forms a square with equal radii's as its sides.
why can i not use 1/2r^2theta now to find the area of sector and area 1/2r^2sintheta of one circle ,subtract them and multiply by two to find the total common area?

4. Hello kandyfloss
Originally Posted by kandyfloss
Yes 11.6 was the radius and the answer is right 118.9.
i tried doing the same thing but by using the angle AOC (pls see the picture)
as 90 deg..because it forms a square with equal radii's as its sides.
why can i not use 1/2r^2theta now to find the area of sector and area 1/2r^2sintheta of one circle ,subtract them and multiply by two to find the total common area?
But it isn't a square just because two adjacent sides are equal, is it? It's simply a rhombus.

If $\displaystyle \angle AOC = 90^o$, then $\displaystyle \theta$ would have to be $\displaystyle 45^o$. Which it isn't. It's $\displaystyle \arccos (7/11.6)=52.88^o$