In triangle ABC, D is the midpoint of BC. Given that AB=3cm, AC=5cm and BC=7cm, find AD.
+9575 We can use Law of Cosines twice.
First, we use it on triangle ABC to find the cosine of angle B.
\(\cos \angle B = \dfrac{3^2 + 7^2 - 5^2}{2(3)(7)}\\ \cos \angle B = \dfrac{11}{14}\)
Then, we use it again, but this time on triangle ABD.
\(AD^2 = AB^2 + BD^2 - 2(AB)(BD) \cos \angle B\\ AD^2 = 3^2 + \left(\dfrac72\right)^2 - 2(3)\left(\dfrac72\right)\left(\dfrac{11}{14}\right)\\ AD^2 = \dfrac{19}4\\ AD = \dfrac{\sqrt{19}}2\text{ cm}\)
