Question

A triangular lot has sides of 215m, 185m, and 125m. Find the measures of the angles at its corners

Answer 1

Answer:

The measures of the angles at its corners are $59.1\°,35.4\°,85.5\°$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of angle A

Applying the law of cosines

$185^{2}= 215^{2}+125^{2}-2(215)(125)cos(A)$

$2(215)(125)cos(A)= 215^{2}+125^{2}-185^{2}$

$cos(A)= [215^{2}+125^{2}-185^{2}]/(2(215)(125))$$cos(A)=0.513953$

$A=arccos(0.513953)=59.1\°$

step 2

Find the measure of angle B

Applying the law of cosines

$125^{2}= 215^{2}+185^{2}-2(215)(185)cos(B)$

$2(215)(185)cos(B)= 215^{2}+185^{2}-125^{2}$

$cos(B)= [215^{2}+185^{2}-125^{2}]/(2(215)(185))$$cos(B)=0.81489$

$B=arccos(0.81489)=35.4\°$

step 3

Find the measure of angle C

Applying the law of cosines

$215^{2}= 125^{2}+185^{2}-2(125)(185)cos(C)$

$2(125)(185)cos(C)= 125^{2}+185^{2}-215^{2}$

$cos(C)= [125^{2}+185^{2}-215^{2}]/(2(125)(185))$$cos(C)=0.0784$

$C=arccos(0.0784)=85.5\°$