You 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.
One side of triangle is 10 centimetres, two angles next to that side are 30° and 60°. I need to find all three angle bisectors of triangle.
3 answers
It 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...
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...
Make 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
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