2007-01-09 04:08:26 · 4 answers · asked by Rich 1 in Science & Mathematics ➔ Mathematics
sin(A-B)=sinAcosB-cosAsinB
sin(A+B)=sinAcosB+cosAsinB
sin(60-x)= sin60cosx-cos60 sinx
=2sinx
sin60cosx=(cos60+2)sinx
tanx= (sin60)/(2+cos60)=a
x1= tan inverse a
x2= 180+tan inverse a.
....................................................................................................
2sin(30+x)=3cosx
2(sin30cosx+cos30sinx)=3cosx
2cos30sinx= (3-2sin30)cosx
tanx= (3-2sin30)/(2cos30)...
you folow the same method as above to get both the roots
2007-01-09 04:22:32 · answer #1 · answered by robin 3 · 0⤊ 0⤋
upload the 8 to the different side. this offers you 15x-3x=24. when you consider that 15x and 3x both have an x in them, you could combine them ensuing in 12x = 24. Now, you purely be certain for x with information from dividing 24 with information from 12. for this reason, x equals 2. examine your paintings, 15(2) - 8 - 3(2) = 16 that's 30 - 8 - 6 = 16....and 16 = 16. you're carried out!
2016-12-02 01:20:11 · answer #2 · answered by huehn 3 · 0⤊ 0⤋
I bet that'd be pretty easy if you know how to arrange sin(α+ß) and then the easy calculations for sin (30º) and sin (60º), and knew that cos (30º) = sin (60º).
2007-01-09 04:12:53 · answer #3 · answered by bequalming 5 · 0⤊ 0⤋
1]
LHS
=sin[60-x]
=sin60*cosx-cos60*sinx
=sqrt3/2*cosx-1/2*sinx
=RHS=2sinx
sqrt3/2*cosx=3/2 sinx
tanx=1/sqrt3
x=30deg
2]
2sin[30+x]=2[1/2*cosx+sqrt3/2*sinx]
=cosx+sqrt3 sinx
=3cosx
sqrt3sinx=2cosx
tanx=2/sqrt3
=1.1547
x=49deg 10min
2007-01-09 04:27:57 · answer #4 · answered by openpsychy 6 · 0⤊ 0⤋