Solve a triangle from one angle and two side lengths

Key concept: This is a triangle solution problem where one angle and two sides are known. Since the given side opposite the known angle is available, the Law of Sines is the cleanest route. After that, a quick angle sum finishes the job.

Step by step

Given:

• $m\angle B=55^\circ40'$
• $AC=8.94\text{ m}$, so $b=8.94$
• $BC=25.1\text{ m}$, so $a=25.1$

We want $\angle A$, $\angle C$, and side $AB=c$.

1) Use the Law of Sines

$$\frac{a}{\sin A}=\frac{b}{\sin B}$$

So

$$\sin A=\frac{a\sin B}{b}$$

Substitute the values:

$$\sin A=\frac{25.1\sin(55^\circ40')}{8.94}$$

Now $\sin(55^\circ40')\approx 0.8256$.

$$\sin A\approx \frac{25.1(0.8256)}{8.94}\approx 2.32$$

That is impossible, since sine values must be between $-1$ and $1$.

2) Check what that means

The given measurements do not form a valid triangle. So there is no triangle with these exact values.

Conclusion

There is no solution.

$\boxed{\text{No triangle exists with the given measurements}}$

Pitfall alert

The biggest trap here is assuming every triangle data set works. Before pushing ahead with Law of Sines, check whether the numbers even make sense. If $\frac{a\sin B}{b}>1$, the triangle cannot exist. That test saves a lot of time.

Try different conditions

If the problem meant a different side labeling, the calculation could change completely. For example, if $AC$ were the side opposite $B$ and $BC$ were not side $a$, then the Law of Sines setup would need to be rebuilt from the diagram. In triangle problems, labels matter as much as the numbers.

Further reading

Law of Sines, triangle inequality, ambiguous data