The failuribe
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Answer:
wtf !! u have written in this question? can u pronounce it??
Answer:
acosA + bsinB = c (1)
asinB - bcosA = k (2)
We need to find k
Lets square both the equations
(1)^{2}(1)
2
---->
(acosA + bsinB)^{2} = c^{2}(acosA+bsinB)
2
=c
2
a^{2} cos^{2}A+b^{2}sin^{2}B+2abcosAsinB = c^{2}a
2
cos
2
A+b
2
sin
2
B+2abcosAsinB=c
2
(3)
(2)^{2}(2)
2
---->
(asinB - bcosA)^{2} = k^{2}(asinB−bcosA)
2
=k
2
a^{2}sin^{2}B+b^{2}cos^{2}A-2absinBcosA = k^{2}a
2
sin
2
B+b
2
cos
2
A−2absinBcosA=k
2
(4)
Add (3) and (4)
a^{2} cos^{2}A+b^{2}sin^{2}B+2abcosAsinBa
2
cos
2
A+b
2
sin
2
B+2abcosAsinB +a^{2}sin^{2}A+b^{2}cos^{2}B-2absinBcosAa
2
sin
2
A+b
2
cos
2
B−2absinBcosA = c^{2}+k^{2}c
2
+k
2
a^{2}[cos^{2}A+sin^{2}A] +b^{2}[sin^{2}B+cos^{2}B]+2abcosAsinB-2absinBcosA = c^{2}+k^{2}a
2
[cos
2
A+sin
2
A]+b
2
[sin
2
B+cos
2
B]+2abcosAsinB−2absinBcosA=c
2
+k
2
a^{2}+b^{2}=c^{2}+k^{2}a
2
+b
2
=c
2
+k
2
k = \sqrt{a^{2}+b^{2}-c^{2}}k=
a
2
+b
2
−c
2
asinB-bcosA = \sqrt{a^{2}+b^{2}-c^{2}}asinB−bcosA=
a
2
+b
2
−c
2