Question

Given that ABC has mA = 63°, b = 9, and c = 11, find the remaining side length a and angles B and C, rounded to the nearest tenth.

Answer 1

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:

`a^2=b^2+c^2-2bc\cos A`

Now, let's substitute the given to the formula.

`\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}`

Now, to find the angles B and C, we will use the sine law.

Sine Law:

`(a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C)`

Now, let's use the value of a in the sine law. Let's get angle B first.

`\begin{gathered} (a)/(\sin A)=(b)/(\sin B) \\ a\sin B=b\sin A \\ (a\sin B)/(a)=(b\sin A)/(a) \\ \sin B=(b\sin A)/(a) \\ B=\sin ^(-1)((b\sin A)/(a)) \\ B=\sin ^(-1)((9\sin63)/(10.58819536)) \\ B=49.2318706 \end{gathered}`

Let's repeat the process to get angle C.

`\begin{gathered} (a)/(\sin A)=(c)/(\sin C) \\ a\sin C=c\sin A \\ (a\sin C)/(a)=(b\sin A)/(a) \\ \sin C=(c\sin A)/(a) \\ C=\sin ^(-1)((c\sin A)/(a)) \\ C=\sin ^(-1)((11\sin 63)/(10.58819536)) \\ C=67.7681294 \end{gathered}`

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.

`\begin{gathered} A+B+C=180 \\ 63+49.2318706+67.7681294=180 \\ 180=180 \end{gathered}`

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.)