SOLUTION: Two angles in a triangle are 49 and 61. If the longest side of the triangle is 15 cm longer than the shortest side, what are the lengths of all three sides.
I was given this que
Algebra ->
Triangles
-> SOLUTION: Two angles in a triangle are 49 and 61. If the longest side of the triangle is 15 cm longer than the shortest side, what are the lengths of all three sides.
I was given this que
Log On
Question 807125: Two angles in a triangle are 49 and 61. If the longest side of the triangle is 15 cm longer than the shortest side, what are the lengths of all three sides.
I was given this question and told the sin law can be applied.
I have tried this:
(x+10)/sin70=x/sin49
x+10(sin49)=x(sin70)
(x+10(sin49))/sin70=x
(x+10(sin49))/sin70=x
but I get stuck here and don't understand how to continue. Help would be appreciated.
Thanks
Spencer Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Two angles in a triangle are 49 and 61. If the longest side of the triangle is 15 cm longer than the shortest side, what are the lengths of all three sides.
I was given this question and told the sin law can be applied.
I have tried this:
(x+10)/sin70=x/sin49
x+10(sin49)=x(sin70)
(x+10(sin49))/sin70=x
(x+10(sin49))/sin70=x
=========================
It says 15 cm longer, why x+10 ?
o/w, you were doing it right.
(x+15)/sin70=x/sin49
(x+15)*sin(49) = x*sin(70)
x*sin(49) + 15sin(49) = xsin(70)
x*sin(70) - x*sin(49) = 15*sin(49)
x*(sin(70) - sin(49)) = 15*sin(49)
x = 15*sin(49)/(sin(70) - sin(49))
Just calculator work from there.