Find the area of a segment of a circle, given radius and central angle. 
Find the Area of a segment of a circle if the central angle of the segment is $105^\circ$ degrees and the radius is $70$.


Formulas I have:

*

*Area of a non-right angle triangle= $\frac{1}{2}a b \sin C$.

*Area of segment = ( area of sector ) $-$ (area of triangle).

Please, could you explain it step by step so I can understand, thanks
 A: You can work out the area of the sector then subtract the area of the triangle.
The area of a sector is given by $\frac{1}{2}r^2\theta$ if $\theta$ is in radians or $\frac{1}{2}r^2\pi\frac{\theta}{180^\circ}$ if $\theta$ is in degrees.
The area of the triangle is give by $\frac{1}{2}r^2\sin\theta$.
Combining these two gives: $\frac{1}{2}\times70^2\times\pi\times\frac{105^\circ}{180^\circ}-\frac{1}{2}\times70^2\times\sin105^\circ \approx 2123.34$
A: Formulas that you need to know:-
(1) The area formula for $\triangle ABC$ $[A = \dfrac 12 ab \sin C]$. (Edited.)
(2) The area of a sector (OAB) formula $[A = \dfrac 12 r^2 \theta]$; where $\theta$ is the central angle and it should be in radian instead of degree. 
Added. The conversion formula is $[\pi$ radians $= 180^0]$. 

Additional info: The common naming convention and the derivation of the area formula:-
We already know that $[⊿ABC] = \dfrac {1}{2}b \times h$.

Since $h = a \sin C$, then $[⊿ABC] = \dfrac {1}{2}ab \times \sin C$.
A: Forget about particular values
and name the things to be measured.
Call the radius of the circle
$r$
and the angle subtended
by the sector $t$.
The area of the part of the circle
containing the sector
and internal triangle
is
$tr^2/2$.
The internal triangle
has altitude
$r\cos(t/2)$
and base
$2r\sin(t/2)$,
so its area is
$\begin{array}\\
(1/2)(2r\sin(t/2))(r\cos(t/2))
&=r^2(\sin(t/2)\cos(t/2))\\
&=r^2\sin(t)/2\\
\end{array}
$
since
$\sin(t)
=2\sin(t/2)\cos(t/2)
$.
The area of the sector
is the difference 
of these expressions,
which is
$tr^2/2-r^2\sin(t)/2
=r^2(t-\sin(t))/2
$.
Now you can substitute
the values of the radius
and angle of the sector.
As a check,
for $t = \pi$,
this gives
$\pi r^2/2$,
which is area of
the semicircle.
