Notes: `A = 110^@, a = 125, b = 200` Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer...

Sunday, 30 November 2014

Given: $\displaystyle{A}={{100}}^{\circ},{a}={125},{b}={200}$

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

$\displaystyle\frac{{125}}{{\sin{{\left({100}\right)}}}}=\frac{{200}}{{\sin{{\left({B}\right)}}}}=\frac{{c}}{{\sin{{\left({C}\right)}}}}$

$\displaystyle\frac{{125}}{{\sin{{\left({100}\right)}}}}=\frac{{200}}{{\sin{{\left({B}\right)}}}}$

$\displaystyle{\sin{{\left({B}\right)}}}=\frac{{{200}{\sin{{\left({100}\right)}}}}}{{125}}$

$\displaystyle{\sin{{\left({B}\right)}}}={1.57}$

$\displaystyle{B}={\arcsin{{\left({1.57}\right)}}}$

It is not possible to find angle B. A triangle cannot be formed from the given information.

Given: $\displaystyle{A}={{100}}^{\circ},{a}={125},{b}={200}$

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

$\displaystyle\frac{{125}}{{\sin{{\left({100}\right)}}}}=\frac{{200}}{{\sin{{\left({B}\right)}}}}=\frac{{c}}{{\sin{{\left({C}\right)}}}}$

$\displaystyle\frac{{125}}{{\sin{{\left({100}\right)}}}}=\frac{{200}}{{\sin{{\left({B}\right)}}}}$

$\displaystyle{\sin{{\left({B}\right)}}}=\frac{{{200}{\sin{{\left({100}\right)}}}}}{{125}}$

$\displaystyle{\sin{{\left({B}\right)}}}={1.57}$

$\displaystyle{B}={\arcsin{{\left({1.57}\right)}}}$

It is not possible to find angle B. A triangle cannot be formed from the given information.

Notes