Given:
angle A = 63 degrees
side b = 9
side c = 11
Asked: Find angles B and C and the length of side a.
Solution:
First, we need to find the length of side a using the cosine law.
Cosine Law:
[tex]a^2=b^2+c^2-2bc\cos A[/tex]
Now, let's substitute the given to the formula.
[tex]\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ a=\sqrt[]{b^2+c^2-2bc\cos A} \\ a=\sqrt[]{9^2+11^2-2\cdot11\cdot9c\cos63} \\ a=10.58819536 \end{gathered}[/tex]
Now, to find the angles B and C, we will use the sine law.
Sine Law:
[tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}[/tex]
Now, let's use the value of a in the sine law. Let's get angle B first.
[tex]\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B} \\ a\sin B=b\sin A \\ \frac{a\sin B}{a}=\frac{b\sin A}{a} \\ \sin B=\frac{b\sin A}{a} \\ B=\sin ^{-1}(\frac{b\sin A}{a}) \\ B=\sin ^{-1}(\frac{9\sin63}{10.58819536}) \\ B=49.2318706 \end{gathered}[/tex]
Let's repeat the process to get angle C.
[tex]\begin{gathered} \frac{a}{\sin A}=\frac{c}{\sin C} \\ a\sin C=c\sin A \\ \frac{a\sin C}{a}=\frac{b\sin A}{a} \\ \sin C=\frac{c\sin A}{a} \\ C=\sin ^{-1}(\frac{c\sin A}{a}) \\ C=\sin ^{-1}(\frac{11\sin 63}{10.58819536}) \\ C=67.7681294 \end{gathered}[/tex]
Note: The sum of the internal angles of a triangle is always 180 degrees.
We can check or work if it is equal to 180 degrees, then everything is correct.
[tex]\begin{gathered} A+B+C=180 \\ 63+49.2318706+67.7681294=180 \\ 180=180 \end{gathered}[/tex]
ANSWER:
length of side a = 10.6 (Rounded to the nearest tenth.)
Angle B = 49.2 degrees (Rounded to the nearest tenth.)
Angle C = 67.8 degrees (Rounded to the nearest tenth.)
