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# The angles $\alpha ,\beta ,\lambda$ of a triangle satisfy the equations $2\sin \alpha + 3\cos \beta = 3\sqrt 2$ and $3\sin \beta + 2\cos \alpha = 1$. Then angle $\lambda$ equals to?

Last updated date: 28th Mar 2023
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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.