Okay, got a few problems I cannot solve. Any help would be great.

1) Find the area of the shaded figure (It's a diamond shaped figure with the top left side 5, top right side 6, bottom left is 8, and bottom right is 7, and the top corner is 100 degrees between the 5 and the 6. Use Law of Cosines and Heron's formula for the area of the triangle.

2) Show that given the three angles A, B, and C of a triangle and one side a (i.e. the side opposite to the angle A) the area of the triangle A is given by the following formula:

A = a(squared)sin(B)sin(C)/2sin(A)

3) Prove the following identity

tan(a-b) + tan(b-c) + tan(c-a) = tan(a-b) tan(b-c) tan(c-a)

2. Hello, ChargersFan84!

These are not easy problems . . .

2) Show that given the three angles $A, B, C$ of a triangle and one side $a,$

the area of the triangle is given by: . $\text{Area} \;=\; \frac{a^2\sin B\sin C}{2\sin A}$
Code:
            B
*
*| *
c * |   *   a
*  |h    *
*   |       *
*    |         *
A * - - * - - - - - * C
D
: - - -  b  - - - :

The area of a triangle is: . $\text{Area} \;=\;\tfrac{1}{2}\text{(base)(height)}\;\;{\color{ blue}[1]}$

In $\Delta ABC$, use the Law of Sines: . $\frac{b}{\sin B} \:=\:\frac{a}{\sin A} \quad\Rightarrow\quad b \:=\:\frac{a\sin B}{\sin A}\;\;\text{(base)}\;\;{\color{blue}[2]}$

In right triangle $BDC\!:\;\sin C \:=\:\frac{h}{a} \quad\Rightarrow\quad h \:=\:a\sin C\;\;\text{(height)}\;\;{\color{blue}[3]}$

Substitute [2] and [3] into [1]: . $\text{Area} \;=\;\frac{1}{2}\left(\frac{a\sin B}{\sin A}\right)(a\sin C) \;=\;\frac{a^2\sin B\sin C}{2\sin A}$

3. ## Haha no they're not.

Thanks for the help on that one. They're not the easiest problems, that's for sure.