The ratios of the lengths of the sides BC and AC of a triangle ABC to the radius of the circumscribed circle are equal to 2 and 3/2 respectively.Find the ratio of the lengths of the bisectors of internal angles of B and C. We are given BC/R=a/R=2 and AC/R=b/R=3/2,where R is the circumradius of the triangle ABC. a=2R,b=3/2 R

Brenda Jordan

Brenda Jordan

Answered question

2022-11-08

The ratios of the lengths of the sides B C and A C of a triangle A B C to the radius of the circumscribed circle are equal to 2 and 3 2 respectively.Find the ratio of the lengths of the bisectors of internal angles of B and C.
We are given B C R = a R = 2 and A C R = b R = 3 2 ,where R is the circumradius of the triangle A B C.
a = 2 R , b = 3 2 R
I know that m b =length of angle bisector of angle B = 2 a c s ( s b ) a + c and m c =length of angle bisector of angle C = 2 a b s ( s c ) a + b but i need the third side in order to use these formulae,which i do not know.What should i do?

Answer & Explanation

Patrick Arnold

Patrick Arnold

Beginner2022-11-09Added 21 answers

1) prove that A = π 2 : Using consequence of sin-theorem :
b = 2 R sin A
Now you could finish by yourself.
Emmanuel Giles

Emmanuel Giles

Beginner2022-11-10Added 2 answers

Use known formula of the bisector length
m c = 2 a b cos ( A / 2 ) a + b
and triangle half angle identity
cos ( C / 2 ) = s ( s c ) a b
to get bisector length in your formula. You can use the Sine Rule and cycle the sides with symmetry.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Elementary geometry

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?