We have triangle $ABC$ with $BC=13,AC=14,hat(C)=arccos(5/13)$ .How to find the other two angles?

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Answer 1 · 2017-04-21T20:20:27

See below.

Using the so called sinus law

$(sin hatA)/[BC] = (sin hat B)/[AC]$

and

$hatA+hatB+hatC=pi$ but $hatC=arccos(5/13)$

and

$sin hatB = sin(pi-hat A-hat C)=sin(hatA+hat C)$ and

$sin(hatA+hat C)=sin hat A cos hat C+cos hat A sin hat C$

and

$sin hat C=12/31$ and $cos hat C = 5/13$

so

$14 sin hat A = 13(5/13 sin hat A+12/13cos hat A)$ or

$9 sin hat A = 12 cos hat A$ and

$tan hat A = 12/9 = 4/3$

and $hat A = arctan(4/3)$

and now

$hat B = pi - (hat A + hat C)$

This is left to the reader.

We have triangle ABCABC with BC=13,AC=14,hat(C)=arccos(5/13)BC=13,AC=14,hat(C)=arccos(5/13).How to find the other two angles?