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.

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calculista calculista · Answer 1 · 2019-12-28T11:21:45+01:00

Answer:

Part a) [tex]AC=135.95\ m[/tex]

Part b) [tex]BC=124.14\ m[/tex]

The diagram in the attached figure

Step-by-step 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

[tex]m\angle A+m\angle B+m\angle C=180^o[/tex]

substitute the given values

[tex]53^o+61^o+m\angle C=180^o[/tex]

[tex]114^o+m\angle C=180^o[/tex]

[tex]m\angle C=180^o-114^o[/tex]

[tex]m\angle C=66^o[/tex]

step 2

Find the distance AC

Applying the law of sines

[tex]\frac{AB}{sin(C)}=\frac{AC}{sin(B)}[/tex]

see the attached figure to better understand the problem

substitute the given values

[tex]\frac{142}{sin(66^o)}=\frac{AC}{sin(61^o)}[/tex]

[tex]AC=\frac{142}{sin(66^o)}(sin(61^o))[/tex]

[tex]AC=135.95\ m[/tex] ---> rounded to the nearest hundredth

step 3

Find the distance BC

Applying the law of sines

[tex]\frac{AB}{sin(C)}=\frac{BC}{sin(A)}[/tex]

substitute the given values

[tex]\frac{142}{sin(66^o)}=\frac{BC}{sin(53^o)}[/tex]

[tex]BC=\frac{142}{sin(66^o)}(sin(53^o))[/tex]

[tex]BC=124.14\ m[/tex] ---> rounded to the nearest hundredth