Question

Solve the triangle.a=9 , c= 6 , B= 99.7 degrees

Answer 1

Given the triangle:

We can use the law of cosines to find the length of side b:

`\begin{gathered} b^2=a^2+c^2-2\cdot a\cdot c\cdot\cos B \\ b^2=9^2+6^2-2\cdot9\cdot6\cdot\cos99.7\degree \\ \therefore b=11.6 \end{gathered}`

Using this value, we can calculate the measure of the angles A and C using the law of sines. For the angle C:

`\begin{gathered} (\sin C)/(c)=(\sin B)/(b) \\ \\ (\sin C)/(6)=(\sin99.7\degree)/(11.6) \\ \\ C=\arcsin((6\sin99.7\degree)/(11.6)) \\ \\ \therefore C=30.7\degree \end{gathered}`

And for angle A:

`\begin{gathered} (\sin A)/(a)=(\sin B)/(b) \\ \\ (\sin A)/(9)=(\sin99.7\degree)/(11.6) \\ \\ A=\arcsin((9\sin99.7\degree)/(11.6)) \\ \\ \therefore A=49.9\degree \end{gathered}`