To solve this problem, we are given a triangle $\triangle LMN$ with the following information:

• $\angle M = 65^\circ$
• $\angle N = 87^\circ$
• Side $n = 13$ mm (which we assume to be opposite $\angle N$).

We are tasked with finding the perimeter of $\triangle LMN$ to the nearest millimeter.

Step 1: Calculate $\angle L$

Since the sum of angles in any triangle is $180^\circ$: $\angle L = 180^\circ - \angle M - \angle N$ $\angle L = 180^\circ - 65^\circ - 87^\circ = 28^\circ$

Step 2: Use the Law of Sines

The Law of Sines states: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ Let:

• Side $m$ be opposite $\angle M = 65^\circ$,
• Side $l$ be opposite $\angle L = 28^\circ$,
• Given $n = 13$ mm is opposite $\angle N = 87^\circ$.

Thus: $\frac{n}{\sin N} = \frac{m}{\sin M} = \frac{l}{\sin L}$

Using $n = 13$ mm and $\angle N = 87^\circ$: $\frac{13}{\sin 87^\circ} = \frac{m}{\sin 65^\circ} = \frac{l}{\sin 28^\circ}$

Step 3: Solve for $m$ and $l$

1. Calculate $m$: $m = \frac{13 \cdot \sin 65^\circ}{\sin 87^\circ}$

2. Calculate $l$: $l = \frac{13 \cdot \sin 28^\circ}{\sin 87^\circ}$

Step 4: Calculate the Perimeter

The perimeter $P$ of the triangle is the sum of all three sides: $P = m + n + l$

Let’s compute this.The perimeter of the triangle $\triangle LMN$ is approximately 31 mm to the nearest millimeter.

Would you like further details, or do you have any questions?

Related Questions

1. How is the Law of Sines applied in different types of triangles?
2. Can the Law of Sines be used in a right triangle? Why or why not?
3. How would the calculation change if we were given two sides and an included angle?
4. What other trigonometric laws are useful for solving triangle problems?
5. How would the solution change if side $m$ was known instead of $n$?

Tip

Remember, the Law of Sines is particularly useful in non-right triangles when you know an angle and its opposite side.