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).
+119 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.
+2864 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.
+118687 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
\(cos30^\circ =\frac{\sqrt3}{2}\\ so\\ \frac{5}{AD}=\frac{\sqrt3}{2}\\ \frac{AD}{5}=\frac{2}{\sqrt3}\\ AD=\frac{10}{\sqrt3}\\ AD=\frac{10\sqrt3}{3}\)
