Question

Calculate the size of $\mathrm{\angle }BAC$ for the following image.

Answer 1

Hint: We will use the trigonometric identity of $\tan \theta$ for the $\mathrm{\angle }BAC$ to find its relation to the sides AB and BC. Then to find the angle we will use the inverse formula of $\tan$.

Complete step by step answer:
For $△ABC$, $\mathrm{\angle }ABC={90}^{\circ }$. So, $△ABC$is a right-angled triangle.
$AB=1.6m,BC=2.8m$.
Let’s assume that $\mathrm{\angle }BAC=\alpha$.
Now, with respect to $\mathrm{\angle }BAC$, we will take the trigonometric identity of $\tan \alpha$.
So, $\tan \alpha ={\displaystyle \frac{height}{base}}={\displaystyle \frac{BC}{AB}}$.
The height and base will be considered with respect to $\mathrm{\angle }BAC$.
Now, we put the values of AB and BC.
So, $\tan \alpha ={\displaystyle \frac{BC}{AB}}={\displaystyle \frac{2.8}{1.6}}={\displaystyle \frac{7}{4}}$.
Thus, from the relation of $\mathrm{\angle }BAC$ to the sides AB and BC we got the value of $\tan \alpha$.
Now, we use the inverse theorem of $\tan$to find the value of the angle.
So,
 $\begin{array}{rl}& \tan \alpha ={\displaystyle \frac{7}{4}}\\ & \Rightarrow \alpha ={\tan }^{-1}\left({\displaystyle \frac{7}{4}}\right)\end{array}$
Thus, we get the value of $\mathrm{\angle }BAC$ as $\alpha ={\tan }^{-1}\left({\displaystyle \frac{7}{4}}\right)$.

Note: Even though we didn’t use the side AC at any point in the solution, still we can do that when we are using the identity of $\sin \theta$ and $\cos \theta$. In that case side AC will be considered as the hypotenuse.
so, $\sin \alpha ={\displaystyle \frac{height}{hypotenuse}}={\displaystyle \frac{BC}{AC}}$ and $\cos \alpha ={\displaystyle \frac{base}{hypotenuse}}={\displaystyle \frac{AB}{AC}}$.
To find the value of side AC we will use Pythagoras’ theorem which tells us that $bas{e}^2+heigh{t}^2=hypotenus{e}^2$
So,
$\begin{array}{rl}& A{B}^2+B{C}^2=A{C}^2\\ & \Rightarrow AC=\sqrt{A{B}^2+B{C}^2}\end{array}$
So, $AC=\sqrt{{\left(1.6\right)}^2+{\left(2.8\right)}^2}=\sqrt{2.56+7.84}=3.22$
Then using the inverse formula, we will find the value of $\mathrm{\angle }BAC$.
Also, we need to remember that the exact solution for $\alpha ={\tan }^{-1}\left({\displaystyle \frac{7}{4}}\right)$ will be taken into consideration not the general solution as $\alpha \in (0,\pi )$.
So, $\alpha ={\tan }^{-1}\left({\displaystyle \frac{7}{4}}\right)={60.25}^{\circ }$(approximation).