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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). 

 

 Dec 1, 2019
 #1
avatar+22 
+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.

 Dec 1, 2019
 #2
avatar+2505 
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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
avatar+105989 
+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

 

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

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 Dec 2, 2019

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