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Question

If $x + y = 75$ and $\sin x : \sin y = \frac{1}{\sqrt{2}} : \frac{1}{2}$, then $x : y$ is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$3:2$

Solving the Trigonometric Ratio $x:y$

We are given two conditions:

  • $x + y = 75$
  • $\sin x : \sin y = \frac{1}{\sqrt{2}} : \frac{1}{2}$

Simplifying the Sine Ratio

First, let's simplify the ratio involving the sines:

$ \frac{\sin x}{\sin y} = \frac{1/\sqrt{2}}{1/2} = \frac{1}{\sqrt{2}} \times \frac{2}{1} = \frac{2}{\sqrt{2}} = \sqrt{2} $

This gives us the equation $\sin x = \sqrt{2} \sin y$.

Finding the Values of $x$ and $y$

Using the sum $x + y = 75$, we can express $y$ as $y = 75 - x$. Substitute this into the simplified sine equation:

$ \sin x = \sqrt{2} \sin(75 - x) $

Expand $\sin(75 - x)$ using the sine subtraction formula ($\sin(A-B) = \sin A \cos B - \cos A \sin B$):

$ \sin x = \sqrt{2} (\sin 75 \cos x - \cos 75 \sin x) $

Rearrange the terms to find $\tan x$:

$ \sin x = \sqrt{2} \sin 75 \cos x - \sqrt{2} \cos 75 \sin x $

$ \sin x (1 + \sqrt{2} \cos 75) = \sqrt{2} \sin 75 \cos x $

$ \frac{\sin x}{\cos x} = \tan x = \frac{\sqrt{2} \sin 75}{1 + \sqrt{2} \cos 75} $

We need the values for $\sin 75^\circ$ and $\cos 75^\circ$. Using angle addition formulas:

  • $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$
  • $\cos 75^\circ = \cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Substitute these values back into the expression for $\tan x$:

$ \tan x = \frac{\sqrt{2} \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right)}{1 + \sqrt{2} \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right)} = \frac{\frac{\sqrt{12} + 2}{4}}{1 + \frac{\sqrt{12} - 2}{4}} = \frac{\frac{2\sqrt{3} + 2}{4}}{\frac{4 + 2\sqrt{3} - 2}{4}} $

$ \tan x = \frac{\frac{\sqrt{3} + 1}{2}}{\frac{2 + 2\sqrt{3}}{4}} = \frac{\frac{\sqrt{3} + 1}{2}}{\frac{1 + \sqrt{3}}{2}} = 1 $

Since $\tan x = 1$, the principal value for $x$ is $45^\circ$.

Now, find $y$ using $x + y = 75^\circ$:

$ y = 75^\circ - x = 75^\circ - 45^\circ = 30^\circ $

Calculating the Ratio $x : y$

The ratio $x : y$ is calculated as:

$ x : y = 45^\circ : 30^\circ $

Simplify the ratio by dividing both sides by their greatest common divisor, which is $15^\circ$:

$ x : y = \frac{45}{15} : \frac{30}{15} = 3 : 2 $

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Important Questions from Trigonometry

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