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

Dec 20, 2016

${m}^{2} = {a}^{2} + {r}^{2} - 2 a r \cos M$

#### Explanation:

Let the triangle be as shown below. Here, we have drawn $R P$ perpendicular to $M A$.

Here we have used Pythogoras theorem in initial calculations.

${m}^{2} = R {P}^{2} + P {A}^{2} = \left(R {M}^{2} - M {P}^{2}\right) + {\left(M A - M P\right)}^{2}$

= $\left(R {M}^{2} - M {P}^{2}\right) + \left(M {A}^{2} - 2 \times M A \times M P + M {P}^{2}\right)$

= ${a}^{2} - M {P}^{2} + {r}^{2} - 2 r \times M P + M {P}^{2}$

= ${a}^{2} - \cancel{M {P}^{2}} + {r}^{2} - 2 r \times M P + \cancel{M {P}^{2}}$

= ${a}^{2} + {r}^{2} - 2 r \times M P$

But as $\cos M = \frac{M P}{R M}$, hence $M P = R M \times \cos M = a \cos M$

Therefore ${m}^{2} = {a}^{2} + {r}^{2} - 2 a r \cos M$