Question

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?

Answer 1

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.