How do you find cos$\left( \theta \right)$ if sin$\left( \theta \right)$ = $\dfrac{3}{5}$ and $90 < \theta < 180$ ?

Hint: In this problem, we will need identities for solving and calculating the value for cos$\left( \theta \right)$ if the value of sin$\left( \theta \right)$ is provided as $\dfrac{3}{5}$. The range implies that in which quadrant the angle will lie. The identities which will be used are:
$\Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x=1$
$\Rightarrow \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$


Complete step by step solution:
Now, let’s discuss the question.
From the given plane, we can say that:
First quadrant consists of the angles which lie in $0 < \theta < 90$. The values of x and y both are positive.
Second quadrant consists of the angles which lie in $90 < \theta < 180$. The values of x are negative whereas y are positive.
Third quadrant consists of the angles which lie in $180 < \theta < 270$. The values of x and y both are negative.
Fourth quadrant consists of the angles which lie in $270 < \theta < 360$. The values of x are positive whereas y are negative.
Let’s see the trigonometric values for sin and cos for different quadrants:


In the given question, the range is given as $90 < \theta < 180$, that means it lies in second quadrant. The value for cos$\left( \theta \right)$ will be negative in this quadrant.
$\Rightarrow $sin$\left( \theta \right)$ = $\dfrac{3}{5}$
As we know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$, obtain the value for cos$\left( \theta \right)$ from this identity:
$\Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Take sin value to the other side:
$\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $
Remove the square of cos and take root of the right hand side:
$\Rightarrow \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Put the negative sign as cos$\left( \theta \right)$ is negative in the second quadrant.
$\Rightarrow \cos \theta =-\sqrt{1-{{\sin }^{2}}\theta }$
Now, place the value of sin$\left( \theta \right)$ which is given:
$\Rightarrow \cos \theta =-\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}$
On solving further:
$\Rightarrow \cos \theta =-\sqrt{1-\dfrac{9}{25}}$
Solve the under root now:
$\Rightarrow \cos \theta =-\sqrt{\dfrac{25-9}{25}}\Leftrightarrow -\sqrt{\dfrac{16}{25}}$
Under root of 16 is 4 and 25 is 5. Remove the under root and obtain the value:
$\therefore \cos \theta =-\dfrac{4}{5}$
This is the final answer.

Note: All the trigonometric identities and functions should be on tips before solving such questions. Also, do remember the square and square roots of the numbers so that there should be no need to find the LCM and then form pairs to check the squares or square root because it consumes a lot of time.