In $\triangle RST$, $RS = 13$, $ST = 14$, and $RT = 15$. Let $M$ be the midpoint of $\overline{ST}$. Find $RM$.
Thanks
We can use the Law of Cosines twice here
RT^2 = ST^2 + RS^2 - 2(ST * RS) cos (RST)
15^2 = 14^2 + 13^2 - 2(14 * 13) cos (RST) simpllfy
225 - 196 - 169 = -364 cos(RST)
-140 = -364cos(RST)
cos RST = 140 / 364 = 5/13
And
RM^2 = SM^2 + RS^2 - 2(SM * RS) cos RST
RM^2 = 7^2 + 13^2 - 2 ( 7 * 13) (5/13) simplify
RM^2 = 49 + 169 - [ 2 *7 * 5 ]
RM^2 = 148
RM = √148 = √ [ 37 * 4 ] = 2√37 ≈ 12.166
Here's another approach using the Law of Cosines only once :
Refer to the following image :
RT^2 = ST^2 + RS^2 - 2(ST * RS) cos (RST)
15^2 = 14^2 + 13^2 - 2(14 * 13) cos (RST) simpllfy
225 - 196 - 169 = -364 cos(RST)
-140 = -364cos(RST)
cos RST = 140 / 364 = 5/13
Now....in triangle RMS draw altitude RN and note that angle RSN = angle RST........since triangle RNS is a right triangle with hypotenuse RS = 13......then.......if cos RST = 5/13...it must be that SN = 5.......so triangle RNS has a hypotenuse of 13 and one leg = 5.....so...the other leg (RN) must = 12
And triangle RMN is another right triangle with RN = 12.....and if M is the midpoint of ST, then SM = 7.......but SN was shown to be = 5....so.....MN = 2
And we have that MN^2 + RN^2 = RM^2 ....so.....
2^2 + 12^2 = RM^2
148 = RM^2
√148 = RM