How do you solve the triangle if b = 12, c=8, a=15?

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Answer 1 · 2018-07-09T11:49:24

$A~~95.08^@, B~~52.83^@, and, C~~32.09^@$ .

Using the cosine formula, we get,

$cosA=(b^2+c^2-a^2)/(2bc)$ ,

$=(12^2+8^2-15^2)/(2*12*8)$ ,

$=(144+64-225)/192$ ,

$=-17/192$ .

$:. A=arccos(-17/192)~~95.08^@$ .

$B=arccos((c^2+a^2-b^2)/(2ca))$ ,

$=arccos{(8^2+15^2-12^2)/(2*8*15)}$ ,

$=arccos(145/240)$ .

$rArr B~~52.83^@$ .

Finally, $C=arccos((a^2+b^2-c^2)/(2ab))$ ,

$=arccos{(15^2+12^2-8^2)/(2*15*12)}$ ,

$=arccos(305/360)$ .

$rArr C~~32.09^@$ .

$or, C=180^@-(95.08^@+52.83^@)~~32.09^@$ .