Let ABC be a triangle with angle B=90 degrees, angle C=30 degrees and AB=5 Find the length of the angle bisector from A to BC (right to 2 decimal places).

Guest Dec 1, 2019

#1**+1 **

I don't know much but it's a 30-60-90 degree triangle and the bisector would bisect A in half and A is 60 so 30? This may be wrong wait for confirmation from somebody else.

OlympusHero Dec 1, 2019

#2**0 **

adding on, now that the bisected angle is 30 degrees, we know that the smaller triangle with the red hypotenuse is a 30-60-90 degree triangle too.

We can calculate the red line now.

CalculatorUser
Dec 1, 2019

#3**+1 **

Let ABC be a triangle with angle B=90 degrees, angle C=30 degrees and AB=5 Find the length of the angle bisector from A to BC (right to 2 decimal places).

Let the intersection of BC and the angle bisector be D

There are differnt ways to do this depending on what topics you have done.

Here is a trigonometry solution but it can be done without trig if this makes no sense to you.

If you do not know trig you can use similar triangles.

triangle ABC is a 30-60-90 triangle just as OlympusHero and calculatorUser have already pointed out.

So

angle BAD=angleDAC=30 degrees.

TRIG SOLUTION

\(cos30^\circ =\frac{\sqrt3}{2}\\ so\\ \frac{5}{AD}=\frac{\sqrt3}{2}\\ \frac{AD}{5}=\frac{2}{\sqrt3}\\ AD=\frac{10}{\sqrt3}\\ AD=\frac{10\sqrt3}{3}\)

.Melody Dec 2, 2019