Question

In ΔMNO, o = 8 inches, mm∠O=48° and mm∠M=28°. Find the length of n, to the nearest 10th of an inch.

Answer 1

Applying the Law of Sines to triangle ΔMNO with known values yields a side length n of approximately 10.35 inches.

Given the triangle ΔMNO, where O is a vertex, and angles O and M are given, along with the side length o opposite to angle O.

To find the length of side n, we can use the Law of Sines:

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

In this case:

o is the side opposite angle O,

n is the side opposite angle N,

m is the side opposite angle M.

We know o = 8 inches, angle O = 48 degrees, and angle M = 28 degrees.

First, find the angle N using the fact that the sum of the angles in a triangle is 180 degrees:

Angle N = 180 degrees - angle O - angle M

Angle N = 180 degrees - 48 degrees - 28 degrees = 104 degrees

Now, apply the Law of Sines:

`\[\frac{o}{\sin(\text{angle } O)} = \frac{n}{\sin(\text{angle } N)}\]`

Substitute the known values:

`\[\frac{8}{\sin(48 \text{ degrees)}} = \frac{n}{\sin(104 \text{ degrees})}\]`

Now, solve for n:

`\[n = \frac{8 \cdot \sin(104 \text{ degrees})}{\sin(48 \text{ degrees})}\]`

Calculate this expression to find the length of n:

`\[n \approx \frac{8 \cdot \sin(104 \text{ degrees})}{\sin(48 \text{ degrees})} \approx 10.35 \text{ inches}\]`

Therefore, the length of n is approximately 10.35 inches (rounded to the nearest tenth of an inch).