Question

If ABC is a triangle with AB=20,BC=22 and CA=24. Let D lie on BC such that AD is the angle bisector of ∠BAC. What is AD^2?

Answer 1

draw a diagram..



find Angle A using the cosine rule

\[\cos(A) = \frac{24^2 + 20^2 - 22^2}{2 \times 24 \times 20}\]

Angle A = 59.1695 degrees.

so the size of 1 bisected angle is 29.5848 degrees

Answer 2

Next find angle ABC using the sin rule

\[\frac{\sin(59.1695)}{22} = \frac{\sin(B)}{24}\]

so \[Sin(B) = \frac{24 \times \sin(59.1695)}{22}\]

so Angle B is 69.5127

then by angle sum of a triangle in triangle ABD

angle BDA = 180 - 59.5127 - 29.5848

so angle BDA = 80.9025

then using the sin rule for triangle ABD

\[\frac{AD}{\sin(69.5127)} = \frac{20}{\sin(80.9025)}\]

solve for AD...

When you get AD just square it for your answer.

hope this helps.,

Answer 3

iitian ? really? :3

Answer 4

I guess there is a quick way using law of cosines. First use the law of cosines on triangle ABC to get an expression for cos(B). Then use the law of cosines again on triangle ABD to get another expression for cos(B). Now equate the two expressions you get and then you can solve for AD^2.