Question

If \( \alpha, \beta \) are the acute angles such that \( \frac{\sin \alpha}{\sin \beta} = \frac{6}{5} \) and \( \frac{\cos \alpha}{\cos \beta} = \frac{9}{5\sqrt{5}} \) then \( \sin \alpha = \)

• A. \( \frac{4}{5} \)
• B. \( \frac{3}{5} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{2}{3} \) Correct Answer

Hint:

When given ratios involving trigonometric functions of two angles (e.g., \( \sin \alpha / \sin \beta \) and \( \cos \alpha / \cos \beta \)), isolate the functions of one angle (e.g., \( \sin \beta \) and \( \cos \beta \)) in terms of the other angle (\( \alpha \)). Then, substitute these into a fundamental trigonometric identity, typically \( \sin^2 \beta + \cos^2 \beta = 1 \), to obtain an equation involving only functions of \( \alpha \). Solve this equation for the required trigonometric function. Remember to consider the quadrant of the angles for signs.

Answer 1

The Correct Option is A

Step 1: Express \( \sin \beta \) and \( \cos \beta \) in terms of \( \sin \alpha \) and \( \cos \alpha \).
From \( \frac{\sin \alpha}{\sin \beta} = \frac{6}{5} \), we have \( \sin \beta = \frac{5}{6} \sin \alpha \).
(Equation 1) From \( \frac{\cos \alpha}{\cos \beta} = \frac{9}{5\sqrt{5}} \), we have \( \cos \beta = \frac{5\sqrt{5}}{9} \cos \alpha \).
(Equation 2) Since \( \alpha \) and \( \beta \) are acute angles, \( \sin \alpha, \cos \alpha, \sin \beta, \cos \beta \) are all positive.

Step 2: Use the trigonometric identity \( \sin^2 \beta + \cos^2 \beta = 1 \).
Substitute the expressions for \( \sin \beta \) and \( \cos \beta \) from Equations 1 and 2: \[ \left(\frac{5}{6} \sin \alpha\right)^2 + \left(\frac{5\sqrt{5}}{9} \cos \alpha\right)^2 = 1 \] \[ \frac{25}{36} \sin^2 \alpha + \frac{25 \cdot 5}{81} \cos^2 \alpha = 1 \] \[ \frac{25}{36} \sin^2 \alpha + \frac{125}{81} \cos^2 \alpha = 1 \]
Step 3: Divide the entire equation by 25.
\[ \frac{1}{36} \sin^2 \alpha + \frac{5}{81} \cos^2 \alpha = \frac{1}{25} \]
Step 4: Substitute \( \cos^2 \alpha = 1 - \sin^2 \alpha \) into the equation.
Let \( s = \sin \alpha \).
Then \( \cos^2 \alpha = 1 - s^2 \).
\[ \frac{1}{36} s^2 + \frac{5}{81} (1 - s^2) = \frac{1}{25} \] \[ \frac{s^2}{36} + \frac{5}{81} - \frac{5s^2}{81} = \frac{1}{25} \] Rearrange terms to solve for \( s^2 \): \[ s^2 \left(\frac{1}{36} - \frac{5}{81}\right) = \frac{1}{25} - \frac{5}{81} \]
Step 5: Calculate the coefficients.
For the left side coefficient: \( \frac{1}{36} - \frac{5}{81} \).
The LCM of 36 and 81 is 324.
\( \frac{1 \cdot 9}{36 \cdot 9} - \frac{5 \cdot 4}{81 \cdot 4} = \frac{9}{324} - \frac{20}{324} = \frac{9-20}{324} = \frac{-11}{324} \).
For the right side: \( \frac{1}{25} - \frac{5}{81} \).
The LCM of 25 and 81 is \( 25 \times 81 = 2025 \).
\( \frac{1 \cdot 81}{25 \cdot 81} - \frac{5 \cdot 25}{81 \cdot 25} = \frac{81}{2025} - \frac{125}{2025} = \frac{81-125}{2025} = \frac{-44}{2025} \).

Step 6: Solve for \( s^2 \).
The equation becomes: \[ s^2 \left(\frac{-11}{324}\right) = \frac{-44}{2025} \] Multiply by -1: \[ s^2 \left(\frac{11}{324}\right) = \frac{44}{2025} \] \[ s^2 = \frac{44}{2025} \cdot \frac{324}{11} \] \[ s^2 = \frac{4 \cdot 11}{2025} \cdot \frac{324}{11} = \frac{4 \cdot 324}{2025} \] Simplify the fraction \( \frac{324}{2025} \): \( 324 = 18^2 = (2 \cdot 3^2)^2 = 4 \cdot 81 \).
\( 2025 = 25 \cdot 81 \).
\[ \frac{324}{2025} = \frac{4 \cdot 81}{25 \cdot 81} = \frac{4}{25} \] So, \( s^2 = 4 \cdot \frac{4}{25} = \frac{16}{25} \).

Step 7: Find \( s = \sin \alpha \).
\[ s = \sqrt{\frac{16}{25}} = \frac{4}{5} \] Since \( \alpha \) is an acute angle, \( \sin \alpha \) must be positive.
So, \( \sin \alpha = \frac{4}{5} \).
This matches option (1).