1. ## trigonometry ABC triangle

of a ABC triangle
a=5 , b=4 and cos(A-B)=31/32

show that c=6

2. Hello, stud_02!

I found a solution . . . but there must be a better way!

In $\displaystyle \Delta ABC\!:\;\;a\:=\:5,\;b\:=\:4,\;\cos(A-B)\:=\:\tfrac{31}{32}$
Show that: .$\displaystyle c\:=\:6$
Code:
              C
*
*| *
* |   *
b=4 *  |     * a=5
*   |y      *
*    |         *
*     |           *
*      |             *
A * - - - * - - - - - - - * B
:   x   D      c-x      :
: - - - - - c - - - - - :

In right triangle $\displaystyle CDA\!:\;\;x^2+y^2\:=\:16 \quad\Rightarrow\quad y^2 \:=\:16-x^2$ .[1]

In right triangle $\displaystyle CDB\!:\;\;(c-x)^2 + y^2 \:=\:25 \quad\Rightarrow\quad y^2 \:=\:25-(c-x)^2$ .[2]

Equate [1] and [2]: .$\displaystyle 16-x^2 \:=\:25 - (c-x)^2 \quad\Rightarrow\quad x \:=\:\frac{c^2-9}{2c}$ .[3]

Substitute into [1]: .$\displaystyle y^2 \:=\:16 - \left(\frac{c^2-9}{2c}\right)^2 \quad\Rightarrow\quad y^2\:=\:\frac{(c^2-1)(81-c^2)}{4c^2}$ .[4]

We are given: .$\displaystyle \cos(A-B) \:=\:\tfrac{31}{32} \quad\Rightarrow\quad \cos A\cos B + \sin A\sin B \:=\:\tfrac{31}{32}$

. . . . $\displaystyle \frac{x}{4}\cdot\frac{c-x}{5} + \frac{y}{4}\cdot\frac{y}{5} \:=\:\frac{31}{32} \quad\Rightarrow\quad x(c-x) + y^2 \:=\:\frac{155}{8}$

Substitute [3] and [4]: .$\displaystyle \left(\frac{c^2-9}{2c}\right)\left(c - \frac{c^2-9}{2c}\right) + \frac{(c^2-1)(81-c^2)}{4c^2} \:=\:\frac{155}{8}$

. . which simplifies to: .$\displaystyle 9c^2 \:=\:324 \quad\Rightarrow\quad c^2 \:=\:36 \quad\Rightarrow\quad\boxed{c \:=\:6}$