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Hint: Use the property $\left( {\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)} \right)$ and $\left( {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1} \right)$.

As you know in triangle, the sum of all the internal angles is equal to $180^\circ $.

$ \Rightarrow \alpha + \beta + \lambda = 180^\circ ................\left( 1 \right)$

Given equations are

$

2\sin \alpha + 3\cos \beta = 3\sqrt 2 \\

3\sin \beta + 2\cos \alpha = 1 \\

$

Now squaring on both sides of the given equations

$

\Rightarrow {\left( {2\sin \alpha + 3\cos \beta } \right)^2} = {\left( {3\sqrt 2 } \right)^2} \\

\Rightarrow {\left( {3\sin \beta + 2\cos \alpha } \right)^2} = {1^2} \\

$

As you know ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$,

$

\Rightarrow {\left( {2\sin \alpha + 3\cos \beta } \right)^2} = {\left( {3\sqrt 2 } \right)^2} \\

= 4{\sin ^2}\alpha + 9{\cos ^2}\beta + 12\sin \alpha \cos \beta = 9 \times 2 = 18.............\left( 2 \right) \\

\Rightarrow {\left( {3\sin \beta + 2\cos \alpha } \right)^2} = {1^2} \\

= 9{\sin ^2}\beta + 4{\cos ^2}\alpha + 12\sin \beta \cos \alpha = 1..................\left( 3 \right) \\

$

Now add equations 2 and 3

$

\Rightarrow 4{\sin ^2}\alpha + 9{\cos ^2}\beta + 12\sin \alpha \cos \beta + 9{\sin ^2}\beta + 4{\cos ^2}\alpha + 12\sin \beta \cos \alpha = 18 + 1 \\

\Rightarrow 9\left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right) + 4\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 12\sin \alpha \cos \beta + 12\sin \beta \cos \alpha = 19 \\

$

Now as we know that, $\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1$

$

\Rightarrow 9\left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right) + 4\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 12\sin \alpha \cos \beta + 12\sin \beta \cos \alpha = 19 \\

= 9 \times 1 + 4 \times 1 + 12\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 19 \\

= 12\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 19 - 9 - 4 = 6 \\

\Rightarrow 2\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 1 \\

$

Now, as we know that $\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)$

$

\Rightarrow 2\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 1 \\

= 2\sin \left( {\alpha + \beta } \right) = 1 \\

= \sin \left( {\alpha + \beta } \right) = \dfrac{1}{2} \\

$

Now from equation 1

$

\alpha + \beta = 180^\circ - \lambda \\

\Rightarrow \sin \left( {\alpha + \beta } \right) = \dfrac{1}{2} \\

\Rightarrow \sin \left( {180^\circ - \lambda } \right) = \dfrac{1}{2} \\

$

Now we know that $\sin \left( {180^\circ - \theta } \right) = \sin \theta $

$ \Rightarrow \sin \lambda = \dfrac{1}{2}$

Now we know that $\dfrac{1}{2}$ is the value of $\sin 30^\circ $

$

\Rightarrow \sin \lambda = \dfrac{1}{2} = \sin 30^\circ \\

\Rightarrow \lambda = 30^\circ \\

$

which is the required value of $\lambda $.

Note: In these types of problems, we should remember that the sum of all internal angles of any triangle is equal to $180^\circ $. We have to modify the equation to get a suitable form such that unnecessary terms can be eliminated when we carry out addition/subtraction. This is followed by the application of the basic trigonometry properties to get the required result.

As you know in triangle, the sum of all the internal angles is equal to $180^\circ $.

$ \Rightarrow \alpha + \beta + \lambda = 180^\circ ................\left( 1 \right)$

Given equations are

$

2\sin \alpha + 3\cos \beta = 3\sqrt 2 \\

3\sin \beta + 2\cos \alpha = 1 \\

$

Now squaring on both sides of the given equations

$

\Rightarrow {\left( {2\sin \alpha + 3\cos \beta } \right)^2} = {\left( {3\sqrt 2 } \right)^2} \\

\Rightarrow {\left( {3\sin \beta + 2\cos \alpha } \right)^2} = {1^2} \\

$

As you know ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$,

$

\Rightarrow {\left( {2\sin \alpha + 3\cos \beta } \right)^2} = {\left( {3\sqrt 2 } \right)^2} \\

= 4{\sin ^2}\alpha + 9{\cos ^2}\beta + 12\sin \alpha \cos \beta = 9 \times 2 = 18.............\left( 2 \right) \\

\Rightarrow {\left( {3\sin \beta + 2\cos \alpha } \right)^2} = {1^2} \\

= 9{\sin ^2}\beta + 4{\cos ^2}\alpha + 12\sin \beta \cos \alpha = 1..................\left( 3 \right) \\

$

Now add equations 2 and 3

$

\Rightarrow 4{\sin ^2}\alpha + 9{\cos ^2}\beta + 12\sin \alpha \cos \beta + 9{\sin ^2}\beta + 4{\cos ^2}\alpha + 12\sin \beta \cos \alpha = 18 + 1 \\

\Rightarrow 9\left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right) + 4\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 12\sin \alpha \cos \beta + 12\sin \beta \cos \alpha = 19 \\

$

Now as we know that, $\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1$

$

\Rightarrow 9\left( {{{\sin }^2}\beta + {{\cos }^2}\beta } \right) + 4\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 12\sin \alpha \cos \beta + 12\sin \beta \cos \alpha = 19 \\

= 9 \times 1 + 4 \times 1 + 12\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 19 \\

= 12\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 19 - 9 - 4 = 6 \\

\Rightarrow 2\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 1 \\

$

Now, as we know that $\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)$

$

\Rightarrow 2\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 1 \\

= 2\sin \left( {\alpha + \beta } \right) = 1 \\

= \sin \left( {\alpha + \beta } \right) = \dfrac{1}{2} \\

$

Now from equation 1

$

\alpha + \beta = 180^\circ - \lambda \\

\Rightarrow \sin \left( {\alpha + \beta } \right) = \dfrac{1}{2} \\

\Rightarrow \sin \left( {180^\circ - \lambda } \right) = \dfrac{1}{2} \\

$

Now we know that $\sin \left( {180^\circ - \theta } \right) = \sin \theta $

$ \Rightarrow \sin \lambda = \dfrac{1}{2}$

Now we know that $\dfrac{1}{2}$ is the value of $\sin 30^\circ $

$

\Rightarrow \sin \lambda = \dfrac{1}{2} = \sin 30^\circ \\

\Rightarrow \lambda = 30^\circ \\

$

which is the required value of $\lambda $.

Note: In these types of problems, we should remember that the sum of all internal angles of any triangle is equal to $180^\circ $. We have to modify the equation to get a suitable form such that unnecessary terms can be eliminated when we carry out addition/subtraction. This is followed by the application of the basic trigonometry properties to get the required result.

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