In Triangle MAR, how do you express #m^2# in terms of a, r, and cosM?

1 Answer
Dec 20, 2016

Answer:

#m^2=a^2+r^2-2arcosM#

Explanation:

Let the triangle be as shown below. Here, we have drawn #RP# perpendicular to #MA#.
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Here we have used Pythogoras theorem in initial calculations.

#m^2=RP^2+PA^2=(RM^2-MP^2)+(MA-MP)^2#

= #(RM^2-MP^2)+(MA^2-2xxMAxxMP+MP^2)#

= #a^2-MP^2+r^2-2rxxMP+MP^2#

= #a^2-cancel(MP^2)+r^2-2rxxMP+cancel(MP^2)#

= #a^2+r^2-2rxxMP#

But as #cosM=(MP)/(RM)#, hence #MP=RMxxcosM=acosM#

Therefore #m^2=a^2+r^2-2arcosM#