Thursday

October 2, 2014

October 2, 2014

Posted by **eve** on Sunday, January 15, 2012 at 9:49am.

- math -
**Reiny**, Sunday, January 15, 2012 at 10:25amYou clearly have a 30-60-90 right-angled triangle where the hypotenuse is 10

So the sides are 5 , 5sqrt(3) and 10.

Did you want to find the lengths of the angle bisectors.

Simple trig calculations would do that.

Give it a shot.

- math -
**eve**, Sunday, January 15, 2012 at 10:52amIt does not. Or I don‘t know what you‘re talking about.

Is these right if you can check, I tried to do it on my own.

10/3√3 ; 10√3/√2+√6 .

didnt get the last one...

- math -
**Reiny**, Sunday, January 15, 2012 at 11:31amMake a sketch, the 30-60-90 triangle has side ratios of 1 : √3 : 2.

Yours is similar to that by a factor of 5

You will have a triangle where the side opposite the 30° angle is 5, the side opposite the 60° angle is 5√3

and the hypotenuse is 10

(check: 5^2 + (5√3)^2 = 10^2 ?, yup!)

Now do one angle bisector at a time:

the 60° bisector, let it be of length x

cos 30° = x/5

x = 5cos 30 = 5(√3/2) = 5√3/2 or appr. 4.33

the 30° bisector, let its length be y

cos15° = y/(5√3)

y = 5√3 cos15 = appr. 8.37

the 90° bisector, let it be z

I will use the Sine Law

z/sin60 = 5/sin75 , (use angle sum of triangle to get 75)

z = 5sin60/sin75 = appr. 4.48

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