Solve "Circles and Sectors: 3θ=2(π−sinθ)

[look416]

Homework Statement

The diagram shows a semicircle APB on AB as diameter. The midpoint of AB is O. The point P on the semicircle is such that the area of the sector POB is equal to twice the area of the shade segment. Given that angle POB is $$\theta$$ radians, show that

3$$\theta$$ = 2($$\pi$$-sin$$\theta$$)​

Homework Equations

The Attempt at a Solution

using formula
Area of circle = $$\frac{1}{2}$$r²$$\theta$$
and
Area of segment = $$\frac{1}{2}$$r² ($$\theta$$ - sin $$\theta$$ )
heres the problems
from the picture http://img130.imageshack.us/img130/1790/001tz.jpg
questions 4
the the angle of the segment is $$\pi$$-$$\theta$$
there I am clueless even i inserted the info i have
what i really get is

$$\theta$$=2[$$\pi$$-$$\theta$$-sin($$\pi$$-$$\theta$$)]​

of course we can't use formula blindly so anyone can help me there

 

[rock.freak667]

If sector POA contains the shaded segment and triangle POA, how do you find the area of the shaded region?

 

[Architeuthis Dux]

I would approach this problem by first analyzing the given information and equations to fully understand the problem at hand. In this case, we are given a semicircle with a point P on its circumference and we are asked to show that 3θ = 2(π-sinθ). This statement involves both angles and areas, so we need to consider both aspects in our solution.

To start, let's consider the area of the sector POB. This can be expressed as ½r2θ, where r is the radius of the semicircle and θ is the central angle of the sector. We also know that the area of the shaded segment is equal to twice the area of the sector. This can be represented as 2(½r2(θ-sinθ)). So, we can set these two expressions equal to each other:

½r2θ = 2(½r2(θ-sinθ))

Simplifying this equation, we get:

r2θ = r2(θ-sinθ)

Now, let's consider the angle POB. We know that it is equal to θ radians. We also know that the angle of the shaded segment is π-θ radians, as stated in the problem. So, we can write the following equation:

θ + (π-θ) = π

Using the trigonometric identity sin(π-x) = sinx, we can rewrite this equation as:

θ + sinθ = π

Substituting this into our original equation, we get:

r2(θ + sinθ) = r2(θ-sinθ)

Expanding this equation, we get:

r2θ + r2sinθ = r2θ - r2sinθ

Simplifying, we get:

r2sinθ = -r2sinθ

Since the left and right sides of the equation are equal, we can conclude that the equation is true for any value of r and θ. Therefore, we have shown that 3θ = 2(π-sinθ).