$\Delta ABC$ is a right triangle, right-angled at C. If A = 30${}^\circ $ and AB = 40 units, find the remaining two sides and $\angle B$ in $\Delta ABC$.

Hint: As, in the $\Delta ABC$ , we are given with angle A and length of the side AB. So, now we can use simple trigonometric ratios of sin and cosine to find the length of the remaining two sides and the angle B of the triangle ABC.

Complete step-by-step answer:


In the above $\Delta ABC$ , concerning the $\angle A$ side BC is the perpendicular, side AC is base and side AB is the hypotenuse respectively.
We know that $sin\theta =\dfrac{perpendicular}{hypotenuse}=\dfrac{BC}{AB}$ .

The angle that we are considering is $\angle A$ . So , $\theta =\angle A=30{}^\circ $ .
Also, we know that $sin30{}^\circ =\dfrac{1}{2}$ .
$\Rightarrow sin\theta =sin30{}^\circ =\dfrac{1}{2}=\dfrac{BC}{AB}.........(i)$

Now, in the question, we are given that the length of side AB is given as AB = 40 units . So, substituting AB = 40 units in equation ( i ), we get:
$\begin{align}
  & \dfrac{BC}{AB}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{BC}{40}=\dfrac{1}{2} \\
 & \Rightarrow BC=\dfrac{40}{2} \\
 & \Rightarrow BC=20 \\
\end{align}$
$\Rightarrow $ Value of BC = 20 units.

Now, we know that in a right-angled triangle, if we know the length of any two sides of the triangle, we can find the third side using the Pythagorean Theorem.

Pythagoras theorem says that in a right-angled triangle, $Hypotenus{{e}^{2}}=Bas{{e}^{2}}+Perpendicula{{r}^{2}}$
Here, the unknown side is AC. So, according to Pythagoras Theorem, $A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}$

Hence, by substituting the values of AB and BC, we get:
${{40}^{2}}=A{{C}^{2}}+{{20}^{2}}$
$\Rightarrow A{{C}^{2}}={{40}^{2}}-{{20}^{2}}$
$\Rightarrow A{{C}^{2}}=1600-400$
$\begin{align}
  & \Rightarrow A{{C}^{2}}=1200 \\
 & \Rightarrow AC=\sqrt{1200} \\
 & \Rightarrow AC=20\sqrt{3} \\
\end{align}$

Therefore, the value of $AC=20\sqrt{3}$units.
Now, we have to find the angle B of the triangle ABC. Let us assume that the angle B = $\theta $ , i.e. $\angle B=\theta $ .

Now, in the above $\Delta ABC$ , concerning the $\angle B$ side AC is the perpendicular, side BC is base and side AB is the hypotenuse respectively.

We know that $sin\theta =\dfrac{perpendicular}{hypotenuse}=\dfrac{AC}{AB}$ .
Here, $AC=20\sqrt{3}$ and AB = 40 units.
$\begin{align}
  & \Rightarrow \sin \theta =\dfrac{20\sqrt{3}}{40}=\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \sin \theta =\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \theta =60{}^\circ \\
\end{align}$

Hence, the value of $\angle B=60{}^\circ $ .

Note: Be careful while substituting the values of the trigonometric functions. Students generally get confused and make silly mistakes.