Question

How do you determine $\sin \theta $ given $\cot \theta =-\dfrac{4}{3},{{270}^{\circ }} < \theta < {{360}^{\circ }}$?

Answer 1

Hint: It is given that $\cot \theta =-\dfrac{4}{3}$ where $\theta $ lies between ${{270}^{\circ }}$ to ${{360}^{\circ }}$ and we are asked to find the value of $\sin \theta $ in that interval so we know that $\cot \theta =\dfrac{B}{P}$ where “B” and “P” is base and the perpendicular of a triangle and then equating this $\cot \theta $ to given $\cot \theta $, we get the value of “P and B” and then using Pythagoras theorem we can find the value of hypotenuse (which we denoted by “H”). Now, we know that $\sin \theta =\dfrac{P}{H}$ and then substitute the value of “P and H” in this formula.

Complete step by step solution:
In the above problem, we have given:
$\cot \theta =-\dfrac{4}{3},{{270}^{\circ }} < \theta < {{360}^{\circ }}$
And the angle $\theta $ lies between ${{270}^{\circ }}\And {{360}^{\circ }}$.
Now, we know that according to trigonometric ratios that $\cot \theta =\dfrac{B}{P}$ where “B” and “P” stands for base and perpendicular in a right angled triangle with respect to angle $\theta $.
In the below diagram, we have shown a right angled triangle ABC:

Now, AB is perpendicular, BC is base and AC is the hypotenuse of the above triangle.
Equating the given $\cot \theta $ and $\cot \theta =\dfrac{B}{P}$ we get,
$\Rightarrow \cot \theta =\dfrac{B}{P}=-\dfrac{4}{3}$
Now, let us remove this negative sign then we get,
$\Rightarrow \cot \theta =\dfrac{B}{P}=\dfrac{4}{3}$
From the above, “B” is equal to 4 and “P” is equal to 3 so we can find the value of hypotenuse (H) by using Pythagoras theorem which is equal to:
$H=\sqrt{{{P}^{2}}+{{B}^{2}}}$
Substituting the values of “P and B” from the above we get,
$\begin{align}
  & H=\sqrt{{{3}^{2}}+{{4}^{2}}} \\
 & \Rightarrow H=\sqrt{9+16} \\
 & \Rightarrow H=\sqrt{25} \\
 & \Rightarrow H=5 \\
\end{align}$
We know that the trigonometric ratio $\sin \theta $ is equal to:
$\sin \theta =\dfrac{P}{H}$
Now, substituting the value of “P and H” in the above equation and we get,
$\Rightarrow \sin \theta =\dfrac{3}{5}$
Now, angle $\theta $ lies between ${{270}^{\circ }}\And {{360}^{\circ }}$ and this is the fourth quadrant and in fourth quadrant sine is negative so putting negative sign in the above we get,
$\Rightarrow \sin \theta =-\dfrac{3}{5}$
Hence, we got the value of $\sin \theta $ as $-\dfrac{3}{5}$.

Note: The mistake that could be possible in the above problem is that you might forget to put negative sign in the value of $\sin \theta $ so make sure to always see the angle given in the question because the range of angle will change the quadrant and hence, will change the sign of the trigonometric ratios.