Welcome to our community

Be a part of something great, join today!

Trigonometry Area/Trigonometry Problem.

Alaba27

New member
Apr 4, 2013
18
Hi again,

I'm completing my tutoring booklet for trigonometry and I have a question that is quite difficult. I've never done anything like it and my solutions page only has the final answer. I'd like to see how the result is found step-by-step, so if anyone can help me out or at least guide me to the solution it would be much appreciated.

http://speedcap.net/sharing/files/57/af/57afa49f650622a931c668f649477a83.png
http://speedcap.net/sharing/files/6a/e5/6ae5a228d615fb949ba020cd1a1c725d.png

If you cannot read the question, it is:

sin(x) = (√3)/2 in the circle with centre O, shown on the right. Furthermore AB // OP and m OA = 1 unit. What is the area of the shaded region?

Thanks in advance!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would suggest finding the area of the circular sector \(\displaystyle OAB\) (you know or can find the radius and the subtended angle), then subtract away the area of \(\displaystyle \Delta OAB\).

I notice you have \(\displaystyle \overline{OA}\) marked as 2, but it should be 1, making \(\displaystyle \overline{AB}=\sqrt{3}\).
 

Alaba27

New member
Apr 4, 2013
18
I would suggest finding the area of the circular sector \(\displaystyle OAB\) (you know or can find the radius and the subtended angle), then subtract away the area of \(\displaystyle \Delta OAB\).

I notice you have \(\displaystyle \overline{OA}\) marked as 2, but it should be 1, making \(\displaystyle \overline{AB}=\sqrt{3}\).
Thanks for the reply! So would OA be identical to OB, then?
 

Alaba27

New member
Apr 4, 2013
18
Alright, I managed to answer my own question. OA and OB are identical. But how can I find the radius of the circle? I guess that is basically what the entire issue is, seeing as how I cannot get the radius from the diagram (I think).

Thanks in advance!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The radius is \(\displaystyle 1=\overline{OA}=\overline{OB}\). Any line segment from the center to a point on the circle is a radius.
 

Alaba27

New member
Apr 4, 2013
18
The radius is \(\displaystyle 1=\overline{OA}=\overline{OB}\). Any line segment from the center to a point on the circle is a radius.
OK, then. But the height of the triangle AOB is still 1/2, am I correct? I have gotten this:

Area of Circle
A = (pi)(r)^2
= (pi)(1)^2
= (pi) units

Area of Triangle AOB
A = (BxH)/2
= (√3 x 0.5)/2
= (√3)/4 units

So I would subtract the Area of the Triangle AOB from the Area of the Circle, and I would get the shaded area?

Thanks in advance!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You want to subtract the area of the triangle (which you have correctly computed) from the area of the circular sector, not the entire circle. Do you see why?
 

Alaba27

New member
Apr 4, 2013
18
You want to subtract the area of the triangle (which you have correctly computed) from the area of the circular sector, not the entire circle. Do you see why?
My apologies, I've never done a question like this before and I don't see anything from my notes or previous questions that resembles something along these lines. So I am going to guess that I have to do this:

> Divide area of entire circle by 2
> Divide segments by 2 both times (because the top right and bottom right segments are broken in half)
> Add together

So, something like this:

Area of entire circle = π units
(The circle has 4 equal segments)
Area of 1 segment = π/4
(2 segments of the same size, so multiply by 2)
Area of both full segments = 0.5π

(For segments that are cut in half, divide π/4 by 2)
Area of half-segment = 1/8π
(Multiply the half-segment by 2 because there are 2 of them)
Area of both half-segments = 0.25π

Add them together = 0.75π

Is this correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Using your method, you could get the correct result, but the picture is misleading you with regards to the angle. The angle is actually larger than it is drawn.

The formula for the area $A_S$ of a circular sector is given by:

\(\displaystyle A_S=\frac{1}{2}r^2\theta\)

Now, we already know $r=1$, and we know \(\displaystyle \theta=2x\), and we know:

\(\displaystyle \sin(x)=\frac{\sqrt{3}}{2}\)

So, can you find $x$, then substitute for $r$ and $\theta$ in the above formula for the area of the sector?
 

Alaba27

New member
Apr 4, 2013
18
Using your method, you could get the correct result, but the picture is misleading you with regards to the angle. The angle is actually larger than it is drawn.

The formula for the area $A_S$ of a circular sector is given by:

\(\displaystyle A_S=\frac{1}{2}r^2\theta\)

Now, we already know $r=1$, and we know \(\displaystyle \theta=2x\), and we know:

\(\displaystyle \sin(x)=\frac{\sqrt{3}}{2}\)

So, can you find $x$, then substitute for $r$ and $\theta$ in the above formula for the area of the sector?
OK, so I followed what you said and got this:

A = (0.5)(r)^2(θ)
= (0.5)(1)^2([√3]/2 x 2)
= (√3)/2 units

Is this correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
No, you are using \(\displaystyle \theta=2\sin(x)\), but what we need is \(\displaystyle \theta=2x\)

If \(\displaystyle \sin(x)=\frac{\sqrt{3}}{2}\) and \(\displaystyle 0<x<\frac{\pi}{2}\) then what is $x$?