1angle C = π/3
plz tell me if iam correct.... i'll post u the solution....
3SINX - 4 SIN3X - K=0
ANGLE A AND ANGLE B SATISFY THE ABOVE EQUATION.
A>B (GIVEN)
FIND ANGLE C?
WHERE A,B AND C ARE THE ANGLES OF A TRIANGLE ABC.......
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7 Answers
11since A and B satisfy the equation,
therefore...... writing the equations in both A and B, we get
3 sin A - 4 sin3A - K = 0 ........ (I)
3 sin B - 4 sin3B - K = 0.........(II)
subtracting (I) from (II), we get
3Sin A - 3Sin B - 4(Sin3A - Sin3B) = 0
3Sin A - 3Sin B - 4(3SinA - Sin 3A - 3SinB + Sin 3B4) = 0
3Sin A - 3Sin B - (3SinA - Sin 3A - 3SinB + Sin 3B) = 0
implies ,
Sin 3A - Sin 3B = 0
2cos(3A+3B2) . sin(3A-3B2) = 0
2cos (3π2 - 3C2) . sin(3A-3B2) = 0 [as A+B+C = π]
- 2cos (3C2) . sin(3A-3B2) = 0
- 2cos (3C2) = 0 ; or sin(3A-3B2) = 0
implies C=Ï€/3 and A=B=Ï€/3
if you 've any problem... u plz solve it urself......
1hey u made two or can be more mistakes here .
1) 2cos(3pie2- 3c2) = -2sin( 3c2) and not -2cos3c2 as u did .....
1u took A=B=PIE/3 ..ISN'T it contradictory becoz in the question it is given that A>B...........
ALSO 3SINA - 4SIN3A=SIN3A ...U MADE IT SO LARGE .....[3]
12cos(3A+3B2).sin(3A-3B2)=0
(3A-3B)/2 CANNOT BE EQUAL TO ZERO AS A>B(GIVEN)
NOW,
(3A+3B)/2=PIE/2
3A + 3B = PIE
A+B = PIE/3
C = PIE - PIE/3
C= 2PIE/3 ......
1oo....sorry....wht a silly mistake i did!!!!..... well.... ugt the ans... gud....