Question

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

Answer 1

Hint: As, in the $\mathrm{\Delta }ABC$ , 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 $\mathrm{\Delta }ABC$ , concerning the $\mathrm{\angle }A$ side BC is the perpendicular, side AC is base and side AB is the hypotenuse respectively.
We know that $sin\theta ={\displaystyle \frac{perpendicular}{hypotenuse}}={\displaystyle \frac{BC}{AB}}$ .

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

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:
$\begin{array}{rl}& {\displaystyle \frac{BC}{AB}}={\displaystyle \frac{1}{2}}\\ & \Rightarrow {\displaystyle \frac{BC}{40}}={\displaystyle \frac{1}{2}}\\ & \Rightarrow BC={\displaystyle \frac{40}{2}}\\ & \Rightarrow BC=20\end{array}$
$\Rightarrow$ 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:
${40}^2=A{C}^2+{20}^2$
$\Rightarrow A{C}^2={40}^2-{20}^2$
$\Rightarrow A{C}^2=1600-400$
$\begin{array}{rl}& \Rightarrow A{C}^2=1200\\ & \Rightarrow AC=\sqrt{1200}\\ & \Rightarrow AC=20\sqrt{3}\end{array}$

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. $\mathrm{\angle }B=\theta$ .

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

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

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

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