Question

In triangle abc,3cosa+4sinb=1 and 3sina+4cosb=6.Then the measure of angle c can be

Answer 1

Please see the attachment

Answer 2

$In \: triangle \: ABC,$

$i ) 3CosA + 4SinB = 1 \:(given)$

/* On squaring both sides , we get */

$\implies (3CosA + 4SinB)^{2} = 1^{2}$

$\implies (3CosA)^{2} + (4SinB)^{2} + 2\times (3CosA )\times (4SinB ) = 1$

$\implies 9Cos^{2}A + 16Sin^{2}B +24CosA sinB ) = 1 \: --(1)$

$ii ) 3Sin A + 4Cos B = 1 \:(given)$

/* On squaring both sides , we get */

$\implies (3SinA + 4CosB)^{2} = 6^{2}$

$\implies (3SinA)^{2} + (4CosB)^{2} + 2\times (3SinA )\times (4Cos B ) = 36$

$\implies 9Sin^{2}A + 16Cos^{2}B +24SinA CosB ) = 36 \: --(2)$

/* Adding equations (1) and (2) , we get */

$9(Cos^{2}A+Sin^{2}A)+16(Sin^{2}B+Cos^{2} A) +24(SinBCosA+SinACosB) = 37$

$\implies 9 + 16 + 24Sin(B+A) = 37$

$\implies 25+ 24Sin(B+A) = 37$

$\implies 24Sin(B+A) = 37 - 25$

$\implies 24Sin(B+A) = 12$

$\implies Sin(B+A) = \frac{12}{24}$

$\implies Sin(B+A) = \frac{1}{2}$

$\implies Sin(B+A) = Sin 30\degree$

$\implies A + B = 30\degree \: --(3)$

$But \: A + B + C = 180\degree \: --(4)$

$\blue {( Angle \:sum \: Property )}$

/* Subtract equation (3) from Equation (4) , we get */

$\angle C = 150\degree$

Therefore.,

$\red { Measure \:of \: \angle C } \green {= 150\degree}$

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