Question

If $\sin A=\dfrac{2}{3}$ , find the value of the other trigonometric ratios.

Answer 1

Hint: We have given the value of sin A so from sin A we can find other trigonometric ratios such as $\cos A,\tan A,\cot A,\sec A\And \cos ecA$ . We know that $\sin A=\dfrac{P}{H}$ from this equation we can find the base of the triangle corresponding to angle A by Pythagora's theorem. Now, we have all the sides so we can easily find the other trigonometric ratios corresponding to angle A.

Complete step-by-step answer:
The value of sin A given in the above question is:
$\sin A=\dfrac{2}{3}$
The below figure is showing a right triangle ABC right angled at B.

In the above figure, “P” stands for perpendicular with respect to angle A, “B” stands for the base of a triangle with respect to angle A and “H” stands for the hypotenuse of the triangle with respect to angle A.
We know from the trigonometric ratio that:
$\sin A=\dfrac{P}{H}$
In the above equation, P stands for perpendicular and H stands for hypotenuse of the triangle so from the Pythagoras theorem we can find the base of the triangle and we are representing a base with a symbol “B”.
$\begin{align}
  & {{H}^{2}}={{P}^{2}}+{{B}^{2}} \\
 & \Rightarrow 9=4+{{B}^{2}} \\
 & \Rightarrow {{B}^{2}}=5 \\
 & \Rightarrow B=\sqrt{5} \\
\end{align}$
Now, we can easily find the other trigonometric ratios with respect to angle A.
$\cos A=\dfrac{B}{H}$
Plugging $B=\sqrt{5}$ and $H=3$ we get,
$\cos A=\dfrac{\sqrt{5}}{3}$
$\tan A=\dfrac{P}{B}$
Plugging the value of $P=2$ and $B=\sqrt{5}$ in the above equation we get,
$\tan A=\dfrac{2}{\sqrt{5}}$
We know that $\cot A$ is the reciprocal of $\tan A$ so,
$\cot A=\dfrac{\sqrt{5}}{2}$
We know that $\cos ecA$ is the reciprocal of $\sin A$ so,
$\cos ecA=\dfrac{3}{2}$
We know that $\sec A$ is the reciprocal of $\cos A$ so,
$\sec A=\dfrac{3}{\sqrt{5}}$
Hence, we have found all the trigonometric ratios corresponding to angle A.

Note: While solving the above problem you might get confused about what are the trigonometric ratios and if you could understand the trigonometric ratios you might get confused like do I have to find the trigonometric ratios for all the angles of the given triangle.
The solution to all this confusion is that trigonometric ratios are the $\cos ,\tan ,\cot ,\sec \And \cos ec$ of a particular angle and as sin A is given in the question so we have to find the trigonometric ratios corresponding to angle A.