Question

In a triangle$ABC$ if $BC=1$ and $AC=2$ , then what is the maximum the value of angle $A$ ?
A. ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)$
B. ${{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$
C. ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)$
D. ${{\tan }^{-1}}\left( \dfrac{1}{3} \right)$

Answer 1

Hint: We will apply sine law to find the maximum value of the given angle for that we will define what a sine law and apply the formula: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ and then find the relation between $\sin A$ and $\sin B$ , then we will find the maximum value of the sin values and then convert into tan and get the answer.

Complete step by step answer:

First, let’s understand what is meant by sine law or law of sines. So, law of sines defines the ratio of sides of a triangle and their respective sine angles are equivalent to each other. The law of sine is used to find the unknown angle or the side of an oblique triangle. The oblique triangle is defined as any triangle, which is not a right triangle. The law of sine should work with at least two angles and its respective side measurements at a time.
In general, the law of sines is defined as the ratio of side length to the sine of the opposite angle. It holds for all the three sides of a triangle respective of their sides and angles.
Sine law = $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$



Now we are given the triangle $ABC$and $BC=1$ and $AC=2$, we will have the following formula:

We will use the sine rule and by that we will have:
$\dfrac{1}{\sin A}=\dfrac{2}{\sin B}\Rightarrow \sin A=\dfrac{\sin B}{2}$
Now, we know that the maximum value of $\sin \theta =1$ and therefore,
The maximum value of $\sin B=1$ which means that the maximum value of $\sin A=\dfrac{1}{2}$
Now , the value of $\cos A=\sqrt{1-{{\sin }^{2}}A}=\sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}=\sqrt{1-\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2}$
As we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ , therefore: $\tan A=\dfrac{\sin A}{\cos A}$ , we will now put the value of $\sin A\text{ and }\cos A$
Therefore: $\tan A=\dfrac{\sin A}{\cos A}\Rightarrow \tan A=\dfrac{\left( \dfrac{1}{2} \right)}{\left( \dfrac{\sqrt{3}}{2} \right)}\Rightarrow \tan A=\dfrac{1}{\sqrt{3}}$
The maximum value of $A={{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$
Hence, the correct option is B.
Note: We found out the maximum value of $\sin \theta =1$ from the range of $\sin \theta $ as it lies between -1 and 1, therefore the minimum value will be $-1$ . Student might make a mistake while applying the sine rule, remember that the numerator the length of opposite side will be written to the angle which is written in the numerator.