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^@#.