Question

A surveyor wants to find the distance from points A and B to an inaccessible point C. These three points form a triangle. Because point C can be sighted from both A and B, he knows that the measure of < A= 53 degrees and the measure of < B = 61 degrees. In addition, the distance from A to B is 142 meters. Find AC and BC. Draw a diagram.

Answer 1

Part a)
`AC=135.95\ m`

Part b)
`BC=124.14\ m`

The diagram in the attached figure

Explanation:

step 1

Find the measure of angle C

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

`m\angle A+m\angle B+m\angle C=180^o`

substitute the given values

`53^o+61^o+m\angle C=180^o`

`114^o+m\angle C=180^o`

`m\angle C=180^o-114^o`

`m\angle C=66^o`

step 2

Find the distance AC

Applying the law of sines

`(AB)/(sin(C))=(AC)/(sin(B))`

see the attached figure to better understand the problem

substitute the given values

`(142)/(sin(66^o))=(AC)/(sin(61^o))`

`AC=(142)/(sin(66^o))(sin(61^o))`

`AC=135.95\ m` ---> rounded to the nearest hundredth

step 3

Find the distance BC

Applying the law of sines

`(AB)/(sin(C))=(BC)/(sin(A))`

substitute the given values

`(142)/(sin(66^o))=(BC)/(sin(53^o))`

`BC=(142)/(sin(66^o))(sin(53^o))`

`BC=124.14\ m` ---> rounded to the nearest hundredth