4. In \( \triangle \mathrm{MNO}, m=820 \mathrm{~cm}, n=360 \mathrm{~cm} \) and \( o=770 \mathrm{~cm} \). Find the measure of \( \angle \mathrm{M} \) to the nearest 10th of a degree.
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To find the measure of angle \( M \) in triangle \( \triangle MNO \), you can use the Law of Cosines, which states: \[ \cos(M) = \frac{n^2 + o^2 - m^2}{2no} \] Plugging in the values \( m = 820 \, \text{cm} \), \( n = 360 \, \text{cm} \), and \( o = 770 \, \text{cm} \): \[ \cos(M) = \frac{360^2 + 770^2 - 820^2}{2 \cdot 360 \cdot 770} \] Calculating the squares: \[ 360^2 = 129600, \quad 770^2 = 592900, \quad 820^2 = 672400 \] Now substituting these values in: \[ \cos(M) = \frac{129600 + 592900 - 672400}{2 \cdot 360 \cdot 770} \] Calculating the numerator: \[ 129600 + 592900 - 672400 = 50000 \] Now, calculating the denominator: \[ 2 \cdot 360 \cdot 770 = 554400 \] So we have: \[ \cos(M) = \frac{50000}{554400} \approx 0.0903 \] Now take the arccosine to find \( M \): \[ M \approx \cos^{-1}(0.0903) \approx 83.8^\circ \] Thus, the measure of \( \angle M \) is approximately \( 83.8^\circ \).
