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Sagot :
The length of the sides,MD, DK, and KM gives the measure of the angle m<D as c) 76°
Which method can be used to find the measure of angle m<D?
In triangle ∆MDK, we have;
m = 16.4
d = 22
k = 19
Using cosine rule, we get;
d² = m² + k² - 2•m•k•cos(D)
Therefore;
2•m•k•cos(D) = m² + k² - d²
[tex]D = arccos \left( \frac{ {m}^{2} + {k}^{2} - {d}^{2} }{2 \cdot m \cdot k} \right)[/tex]
Which gives;
[tex]D = arccos \left( \frac{ {16.4}^{2} + {19}^{2} - {22}^{2} }{2 \times 16.4 \times 19} \right) \approx {76}^{ \circ} [/tex]
- m<D = 76°
The correct option is therefore;
- c) 76°
Learn more about the rule of cosines here:
https://brainly.com/question/4316134
#SPJ1
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