Question

In any triangle ABC, if a=18,b=24,c=30,then find
sinA, sinB and sinC.

Answer 1

law of cosines
then use law of sines

Answer 2

how?

Answer 3

\[c^2=a^2+b^2-2ab \cos(C)\]
\[\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\]

Answer 4

the triangles sides given by u is a pythogarian triplet as 18^2 +24^2=324+576=900=30^2
thus <C =90 thus sinC=1
sinA= 18/30 =3/5
and sinB=24/30=4/5

Answer 5

i want the sol. by law of sines and cosines

Answer 6

@completeidiot continue your sol.

Answer 7

nothing to complete
the first equation is law of cosines, the 2nd is law of sines
use those equations

Answer 8

actually sinA/a=sinC/c bcoz C=90 hence sinA=a sinC /c=18*sin90/30=18/30=3/5
and same for sinB

Answer 9

sinA=1/2 and sinB=2 and sin C=1/2

Answer 10

AB=OB-OC=b-a same for BC and AC

Answer 11

Then we get AB=6 BC=6 AC=12 sinA=6/12 and SinB=12/6 and sinC=6/12