Suppose in a right angle triangle $\cos \left( t \right)=\dfrac{3}{4}$. How do you find: $\sin \left( t \right)$?

Hint: Since $\cos \left( t \right)=\dfrac{b}{h}$, hence $\cos \left( t \right)$ is given means the ratio of $\dfrac{b}{h}$ is given. So first try to find ‘p’ using Pythagoras' theorem i.e. ${{h}^{2}}={{p}^{2}}+{{b}^{2}}$. After finding ‘p’, for $\sin \left( t \right)$ obtain the ratio of $\dfrac{p}{h}$ to get the required solution.

Complete step by step answer:
We know if in a right angle triangle the base is ‘b’, perpendicular is ‘p’ and hypotenuse is ‘h’; then
$\begin{align}
  & \sin \alpha =\dfrac{p}{h} \\
 & \cos \alpha =\dfrac{b}{h} \\
 & \tan \alpha =\dfrac{p}{b} \\
\end{align}$
Similarly $\cot \alpha $,$\sec \alpha $, $\cos ec\alpha $ are the reciprocals of $\sin \alpha $,$\cos \alpha $and $\tan \alpha $ respectively.
Pythagoras' theorem: In a right angle triangle the square of the hypotenuse is equal to the summation of the square of the base and perpendicular. For the above triangle, Pythagoras' theorem can be applied as ${{h}^{2}}={{p}^{2}}+{{b}^{2}}$
Now considering our question
We have $\cos \left( t \right)=\dfrac{3}{4}$
 $\Rightarrow \dfrac{b}{h}=\dfrac{3}{4}$
Using Pythagoras' theorem, we get
$\begin{align}
  & {{h}^{2}}={{p}^{2}}+{{b}^{2}} \\
 & \Rightarrow {{4}^{2}}={{p}^{2}}+{{3}^{2}} \\
 & \Rightarrow 16={{p}^{2}}+9 \\
 & \Rightarrow {{p}^{2}}=16-9 \\
 & \Rightarrow {{p}^{2}}=7 \\
 & \Rightarrow p=\sqrt{7} \\
\end{align}$
As we know $\sin \alpha =\dfrac{p}{h}$
So, $\sin \left( t \right)=\dfrac{\sqrt{7}}{4}$
This is the required solution of the given question.

Note: Since ‘t’ is an angle of a right triangle, it must be acute (as the inner angles of a triangle sum to ${{180}^{\circ }}$, but since one angle is ${{90}^{\circ }}$, the sum of other two must be ${{90}^{\circ }}$. An acute angle lies in the first quadrant, where both sine and cosine are positive. For solving such a question our aim should be to find ‘p’, ‘b’ and ‘h’, so that we can find any trigonometric function. The above question can also be solved directly using the trigonometry formula ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
We have $\cos \left( t \right)=\dfrac{3}{4}$
Putting the value of $\cos \left( t \right)$ in the equation ${{\sin }^{2}}\left( t \right)+{{\cos }^{2}}\left( t \right)=1$
$\begin{align}
  & \Rightarrow {{\sin }^{2}}\left( t \right)+{{\left( \dfrac{3}{4} \right)}^{2}}=1 \\
 & \Rightarrow {{\sin }^{2}}\left( t \right)+\dfrac{9}{16}=1 \\
 & \Rightarrow {{\sin }^{2}}\left( t \right)=1-\dfrac{9}{16} \\
 & \Rightarrow {{\sin }^{2}}\left( t \right)=\dfrac{16-9}{16} \\
 & \Rightarrow {{\sin }^{2}}\left( t \right)=\dfrac{7}{16} \\
 & \Rightarrow \sin \left( t \right)=\sqrt{\dfrac{7}{16}} \\
 & \Rightarrow \sin \left( t \right)=\dfrac{\sqrt{7}}{4} \\
\end{align}$
This is the alternative method.