In #triangle MNP, m=16, n=15, p=8#, how do you find the cosine of each of the angles?

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Answer 1 · 2016-12-09T16:22:09

#cos hatM=0.1375#
#cos hatN=0.37109375#
#cos hatP=0,86875#

If you know two sides and the included angle of a triangle ABC, you can apply the cosine thorem:

#a^2=b^2+c^2-2bc cos hatA#

Now you can use the inverse formula to know the cosine of an angle when you know the three sides:

#cos hatA=(b^2+c^2-a^2)/(2bc)#

Then

#cos hatM=(n^2+p^2-m^2)/(2np)#

#=(15^2+8^2-16^2)/(2*15*8)#

#=0.1375#

#hatM=82.1°#

#cos hatN=(m^2+p^2-n^2)/(2mp)#

#=(16^2+8^2-15^2)/(2*16*8)#

#=0.37109375#

#hatN=68.22°#

#cos hatP=(m^2+n^2-p^2)/(2mn)#

#=(16^2+15^2-8^2)/(2*16*15)#

#=0,86875#

#hatP=29.69°#