Question

From the following figure find the value of $\sin 30{}^\circ ,\cos 30{}^\circ \ and\ \tan 30{}^\circ $.

Answer 1

Hint: We will calculated the value of $\cos \theta =\dfrac{Base}{hypotenuse}$, $\sin \theta =\dfrac{Perpendicular}{hypotenuse}$and $\tan \theta =\dfrac{Perpendicular}{Base}$. All the value of Base, Perpendicular height and hypotenuse is given. So, just put the value in their respective place to get the value of angles.

Complete step-by-step answer:
It is given in the question that PQR is a right angle triangle which is right angled at Q. Now, we will try to find out the value of $\sin 30{}^\circ ,\cos 30{}^\circ \ and\ \tan 30{}^\circ $ respectively.
This right angled triangle PQR is showing the given conditions in question.

Now, we know that in this right angled triangle PQR, there are some specific names of each side as PQ is called height of the triangle, QR is called the base of the triangle, and PR is called the hypotenuse of the triangle PQR.
Also, It is given that PQ or perpendicular height of the triangle is of 5 units, QR or the base of the given triangle is of $5\sqrt{3}$unit and the side PR or hypotenuse of the triangle is of 10 unit.
We know that from Pythagoras Theorem can be applied to a right triangle. So, we have the theorem as,
${{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}={{\left( Hypotenuse \right)}^{2}}$
Also, we have Base = $5\sqrt{3}$, Perpendicular = 5 and Hypotenuse = 10. So, substituting these values, we get,
\[\begin{align}
  & \Rightarrow {{\left( 5\sqrt{3} \right)}^{2}}+{{\left( 5 \right)}^{2}}={{\left( 10 \right)}^{2}} \\
 & \Rightarrow 75+25=100 \\
 & \Rightarrow 100=100 \\
\end{align}\]
Thus, LHS = RHS
Now, we know that $\sin \theta =\dfrac{Perpendicular}{hypotenuse}$. Here from the question, we have perpendicular = 5 and hypotenuse = 10. So, we get,
$\begin{align}
  & \sin \theta =\dfrac{5}{10} \\
 & \sin \theta =\dfrac{1}{2} \\
\end{align}$
Also, we have here $\theta =30{}^\circ $.
$\Rightarrow \sin 30{}^\circ =\dfrac{1}{2}$
Similarly, now we will calculate $\cos \theta $. We know that $\cos \theta =\dfrac{Base}{hypotenuse}$. Here from the question we get Base = $5\sqrt{3}$ and Hypotenuse = 10. So, on putting the value of Base as $5\sqrt{3}$ and Hypotenuse as 10, we get,
$\begin{align}
  & \cos \theta =\dfrac{5\sqrt{3}}{10} \\
 & \Rightarrow \cos \theta =\dfrac{\sqrt{3}}{2} \\
\end{align}$
Also, we have here $\theta =30{}^\circ $. Thus, we get,
$\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$
Now, we will calculate the value of $\tan \theta $. Here, $\theta =30{}^\circ $and we know that $\tan \theta =\dfrac{Perpendicular}{Base}$. Here from the question we get Base = $5\sqrt{3}$ and Perpendicular = 5. So, Putting the value of base as $5\sqrt{3}$ and Perpendicular as 5, we get,
\[\begin{align}
  & \tan \theta =\dfrac{Perpendicular}{Base} \\
 & \tan \theta =\dfrac{5}{5\sqrt{3}} \\
 & \tan \theta =\dfrac{1}{\sqrt{3}} \\
\end{align}\]
Here, we know that $\theta =30{}^\circ $and thus we get,
$\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$
Thus, in triangle PQR, we get $\sin 30{}^\circ =\dfrac{1}{2},\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}\ and\ \tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Note: If the angle and one side is given then also we can calculate all the values $\sin \theta ,\cos \theta ,\tan \theta $. But you need to remember some standard angle value. For example in the given question we have Hypotenuse as 10 as $\cos 30{}^\circ $. So, we can calculate the base value as,
$\begin{align}
  & \cos 30{}^\circ =\dfrac{B}{H} \\
 & \dfrac{\sqrt{3}}{2}=\dfrac{B}{10} \\
 & B=\dfrac{10\sqrt{3}}{2}=5\sqrt{3} \\
\end{align}$
So, using this method we can calculate all the values required in a triangle. We can also calculate value of $\tan \theta \ as\ \dfrac{\sin \theta }{\cos \theta }$.For example, in the given question we have calculated value of $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$ using sides of the given triangle. Now, using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, we get,
$\begin{align}
  & \Rightarrow \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} \\
 & \Rightarrow \dfrac{1}{2}\times \dfrac{2}{\sqrt{3}}=\dfrac{1}{\sqrt{3}} \\
\end{align}$
Thus, from this method also we get $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.