Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Pranto****Member**- Registered: 2013-07-19
- Posts: 1

In ^ABC, AD bisects <ABC and meets BC at D . E is a point on AC sothat EC=1 ,AB=6 ,BD=2 ,CD=3 & DE=(a/root b) ,where a and b are integers, b is a prime .Find a+b .

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi Pranto

Welcome to the forum.

I think you meant AD bisects angle BAC otherwise the diagram won't make sense. (note the angle is at the middle letter)

I have made a diagram. It is not shown accurately to scale.

angle BAD = angle DAC = x

angle BDA = y (note. angle ADC = 180 - y and so these two angles will have the same sine)

angle ACB = z

Use the sine rule on triangle BAD to get an equation involving sin(x) and sin(y)

Use the sine rule again on triangle DAC to get another equation involving the same.

By dividing one by the other you can work out AC.

use the cosine rule on triangle ABC to get an equation involving cos(z)

use the cosine rule on triangle CDE to get an equation involving cos(z) and DE.

Eliminate cos(z) to get DE.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**