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Question
Find the equation of the straight line which passes through the point (1,2) and makes such an angle with the positive direction of x-axis whose sine is \[\frac{3}{5}\].
Solution
Let \[\theta\] be the inclination of the line with the positive x-axis.
Then, we have,
\[\sin\theta = \frac{3}{5}\]
\[ \Rightarrow \tan\theta = \frac{\sin\theta}{\sqrt{1 - \sin^2 \theta}} = \frac{\frac{3}{5}}{\sqrt{1 - \frac{3^2}{5^2}}}\frac{3}{\sqrt{5^2 - 3^2}} = \frac{3}{4}\]
So, the equation of the line that passes through (1, 2) and has slope \[\frac{3}{4}\] is
\[y - 2 = \frac{3}{4}\left( x - 1 \right)\]
\[ \Rightarrow 3x - 4y + 5 = 0\]
Hence, the equation of the required line is \[3x - 4y + 5 = 0\]
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