Question

Solve the triangle. a = 5.10 m, b = 8.73 m, C = 108.5°

Answer 1

The Solution:

Represent the problem in a diagram:

Required:

To find the values of:

`\begin{gathered} \angle A \\ \angle B \\ c \end{gathered}`

Step 1:

Use the Law of cosines to find the value of c.

`\begin{gathered} c^2=a^2+b^2-2ab\cos C \\ \text{ where} \\ c=? \\ a=5.10m \\ b=8.73m \\ C=108.5^o \end{gathered}`

Substituting these values, we get

`\begin{gathered} c^2=5.10^2+8.73^2-(2*5.10*8.73*\cos108.5) \\ \\ c^2=130.47761 \\ \\ c=√(130.47761)=11.4227\approx11.42m \end{gathered}`

Step 2:

Use the Law of sines to find angle A.

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

Substituting, we get:

`\begin{gathered} (\sin A)/(5.10)=(\sin108.5)/(11.42) \\ So, \\ \\ \sin A=(5.10*\sin108.5)/(11.42)=0.423507 \\ \\ A=\sin^(-1)(0.423507)=25.0562\approx25.1^o \end{gathered}`

Step 3:

Find angle B.

By the sum of angles in a triangle:

`\begin{gathered} \angle A+\angle B+\angle C=180^o \\ 25.1+\angle B+108.5=180 \\ \angle B=180-(108.5+25.1) \\ \angle B=180-133.6=46.4^o \end{gathered}`

Therefore, the correct answers are:

`\begin{gathered} \angle A=25.1^o \\ \\ \angle B=46.4^o \\ \\ c=11.42m \end{gathered}`