The sides of a triangle are 15, 20 and 28. How long are the segments into which the bisector of the largest angle separates the opposite side
3 answers
they are in the ratio 3:4 as provided by the angle bisector theorem.
The largest angle will be opposite the side 28
let it be 2Ø, so each bisected angle is Ø
let the bisector form angles A and B along the 28 side, so that A + B = 180° --> B = 180-A
and we know sinA = sin(180-A) = sinB
let the 28 side be split into x and 28-x, where x is adjacent the side 20
Now use the sine law in each of the smaller triangles
sinØ/x = sinA/20
sinØ = x sinA/20
sinØ/(28-x) = sinB/15
sinØ = (28-x)sinB/15
thus:
x sinA/20 = (28-x)sinB/15 , but remember sinA = sinB, so dividing them out
x/20 = (28-x)/15
15x = 560 - 20x
35x = 560
x = 16
so the side 28 is cut into parts 16 and 12
let it be 2Ø, so each bisected angle is Ø
let the bisector form angles A and B along the 28 side, so that A + B = 180° --> B = 180-A
and we know sinA = sin(180-A) = sinB
let the 28 side be split into x and 28-x, where x is adjacent the side 20
Now use the sine law in each of the smaller triangles
sinØ/x = sinA/20
sinØ = x sinA/20
sinØ/(28-x) = sinB/15
sinØ = (28-x)sinB/15
thus:
x sinA/20 = (28-x)sinB/15 , but remember sinA = sinB, so dividing them out
x/20 = (28-x)/15
15x = 560 - 20x
35x = 560
x = 16
so the side 28 is cut into parts 16 and 12
good call Steve!
(at least I got the right answer, lol)
(at least I got the right answer, lol)