Notes: `A = 76^@, a = 18, b = 20` Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer to...

Tuesday, 20 August 2013

Given: $\displaystyle{A}={{76}}^{\circ},{a}={18},{b}={20}$

The Law of Sines: $\displaystyle\frac{{a}}{{\sin{{\left({A}\right)}}}}=\frac{{b}}{{\sin{{\left({B}\right)}}}}=\frac{{c}}{{\sin{{\left({C}\right)}}}}$

$\displaystyle\frac{{18}}{{\sin{{\left({76}\right)}}}}=\frac{{20}}{{\sin{{\left({B}\right)}}}}=\frac{{c}}{{\sin{{\left({C}\right)}}}}$

$\displaystyle\frac{{18}}{{\sin{{\left({76}\right)}}}}=\frac{{20}}{{\sin{{\left({B}\right)}}}}$

$\displaystyle{\sin{{\left({B}\right)}}}=\frac{{{20}{\sin{{\left({76}\right)}}}}}{{18}}$

$\displaystyle{\sin{{\left({B}\right)}}}={1.0781}$

$\displaystyle{B}={\arcsin{{\left({1.0781}\right)}}}$

Angle B does not exist. A triangle cannot be formed using the given information.

Given: $\displaystyle{A}={{76}}^{\circ},{a}={18},{b}={20}$

The Law of Sines: $\displaystyle\frac{{a}}{{\sin{{\left({A}\right)}}}}=\frac{{b}}{{\sin{{\left({B}\right)}}}}=\frac{{c}}{{\sin{{\left({C}\right)}}}}$

$\displaystyle\frac{{18}}{{\sin{{\left({76}\right)}}}}=\frac{{20}}{{\sin{{\left({B}\right)}}}}=\frac{{c}}{{\sin{{\left({C}\right)}}}}$

$\displaystyle\frac{{18}}{{\sin{{\left({76}\right)}}}}=\frac{{20}}{{\sin{{\left({B}\right)}}}}$

$\displaystyle{\sin{{\left({B}\right)}}}=\frac{{{20}{\sin{{\left({76}\right)}}}}}{{18}}$

$\displaystyle{\sin{{\left({B}\right)}}}={1.0781}$

$\displaystyle{B}={\arcsin{{\left({1.0781}\right)}}}$

Angle B does not exist. A triangle cannot be formed using the given information.

Notes