Question

$$\left. \begin{array} { l } { a ^ { 2 } = b ^ { 2 } + c ^ { 2 } } \\ { - 2 b c c \cos A } \end{array} \right.$$

Answer

$$a=sqrt(b*(b+Pr*e^4*sin(e)*c^2*o^2*v*t^2*h^2*r*m*y)),-sqrt(b*(b+Pr*e^4*sin(e)*c^2*o^2*v*t^2*h^2*r*m*y))$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{a}^{2}={b}^{2}+{c}^{2}Pr{o}^{2}v{e}^{4}{t}^{2}{h}^{2}rmby\sin{e}\]
Regroup terms.
\[{a}^{2}={b}^{2}+Pr{e}^{4}\sin{e}{c}^{2}{o}^{2}v{t}^{2}{h}^{2}rmby\]
Factor out the common term \(b\).
\[{a}^{2}=b(b+Pr{e}^{4}\sin{e}{c}^{2}{o}^{2}v{t}^{2}{h}^{2}rmy)\]
Take the square root of both sides.
\[a=\pm \sqrt{b(b+Pr{e}^{4}\sin{e}{c}^{2}{o}^{2}v{t}^{2}{h}^{2}rmy)}\]