ABC is an isosceles triangle with sides AB=AC=5. D is a point in between B and C such that BD=2 and DC=4.5. Find the length of AD.
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This being an isoscles triangle the base angles B and C are equal in measure. Let's call that measure \(\alpha\). Also let's call the length of Ad, x. By the law of cosines, we have
\(x^2=5^2+2^2-2(2)(5)cos(\alpha)\), and \(x^2=5^2+4.5^2-2(5)(4.5)cos(\alpha)\). By setting the two equations equal to each other (i.e. eliminating x squared) we get \(29-20cos(\alpha)=45.25-45cos(\alpha)\). Solving for \(cos(\alpha)\) we get \(cos(\alpha)=0.65\). So
\(x^2=29-20(0.65)=16\), and as result x=4. Good luck and let me know for future reference if I am overdoing the 'help'! Perhaps a couple of hints would have been more than adequate.
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