# Trigonometry using parallel lines

Question

In the following figure,what will be the value of !!(AD)?!!

Solution:

We are given that the perimeter's of !!triangleADB!! and !!triangleADC!! are equal.

The first step would be to find the length of !!BD!! and !!DC!!.

Since the perimiters are equal we get !!AB+BD+AD=AD+DC+CA!!

!!=> AB+BD=AC+DC!!

!!=> 60+100-DC=80+DC!!

!!=> DC=40!! so !!BD = 60!!

Now there are two ways to solve this question. First is a simple one which requires some creativity or construction and basic knowledge of parallel lines.

Let us draw a line !!PD!! parallel to !!AC!!, so we will get two right angled triangles !!BPD!! and !!PDA!!. Since !!PD!! is parallel to !!AC!!

!!((BD)/(BC))=((BP)/(BA))!!

That is, !!(60/100)=((BP)/60)!!

Therefore, !!BP!! !!=!! !!36!!

Now in !!triangle!! !!BDP!!,applying pythagoreas theorem, we get,

!!(PD^2) = (BD^2)-(BP^2)!!

!!PD=sqrt[60^2) - (36^2)]!!

!!PD=48!!

Again, in !!triangle!! !!PDA!!,applying pythagoreas theorem, we get,

!!(AD^2)=PD^2+AP^2!!

!!AD=sqrt[(48^2) + (24^2)]!!

!!AD=sqrt(2880)!!

Therefore, !!AD=24sqrt(5)!!

Alternatively a more mudane approach would be to look at the triangle !!ABD!!. Since we know two sides and an angle we should be able to apply the formula.

!!AD^2=BD^2+AB^2-2BDxxABcosABD!!

Solving this we will get !!AD = 24sqrt(5)!!