Asked by Gems
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
Answers
Answered by
Steve
they are in the ratio 3:4 as provided by the angle bisector theorem.
Answered by
Reiny
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
Answered by
Reiny
good call Steve!
(at least I got the right answer, lol)
(at least I got the right answer, lol)
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