In an equilateral triangle ABC of side 10cm, the side BC is trisected at D. Then the length (in cm) of AD is
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 = $$\frac{10\sqrt7}{3}$$
Create a FREE account and get:
Quick, Easy and Effective Revision
By proceeding you agree to create your account
Free CAT Formulae PDF will be sent to your email address soon !!!
