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Question 129
In an equilateral triangle ABC of side 10cm, the side BC is trisected at D. Then the length (in cm) of AD is
Solution
As we know $$c^2 = a^2 +b^2 - 2ab \times cosC$$ Â ( where $$c$$ is length of AD, $$a$$ is length of AB = 10 and $$b$$ is length of one of trisected part = $$\frac{10}{3}$$)
after putting values and solving it , we will get its value = $$\frac{10\sqrt7}{3}$$
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