Question

If $A={{60}^{0}},a=5,b=4\sqrt{3}$ in triangle$ABC$, then $B=$
A. ${{30}^{0}}$
B. ${{60}^{0}}$
C. ${{90}^{0}}$
D. None of these.

Answer 1

Hint: To solve this question, we will use Law of sine $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$. We will substitute the given values of sides and angles in the first two components of the Law of sine and determine the value of the angle $B$.

Complete step by step solution: We are given a triangle $ABC$ having angle $A={{60}^{0}}$ and value of the length of the sides as $a=5,b=4\sqrt{3}$ and we have to find the value of the angle $B$.
We will use Law of sine $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ and take only the first two component of this relationship that is$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$.
We will now substitute the values of the sides $a=5,b=4\sqrt{3}$and angle $A={{60}^{0}}$in $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$.
$\begin{align}
  & \dfrac{5}{\sin {{60}^{0}}}=\dfrac{4\sqrt{3}}{\sin B} \\
 & \dfrac{5}{\dfrac{\sqrt{3}}{2}}=\dfrac{4\sqrt{3}}{\sin B} \\
 & \sin B=\dfrac{6}{5}
\end{align}$
The value we are getting of sine is greater than one that is $\dfrac{6}{5}>1$. And we know that the value of the trigonometric function sine lies in the interval of $\left[ -1,1 \right]$. Hence the correct option will be none of these.

The value of the angle B of triangle $ABC$ having angle $A={{60}^{0}}$ and value of the length of the sides as $a=5,b=4\sqrt{3}$ is not determined. Hence the correct option is (D).

Note: The trigonometric function sine of any angle is the ratio of the perpendicular to the hypotenuse of a right angled triangle that is $\sin \theta =\dfrac{P}{H}$. The value of this ratio cannot be greater than one.
As we know that the largest side of a right angled triangle is hypotenuse and $\sin \theta =\dfrac{P}{H}$ which means that any value of the perpendicular will be divided by a large number of hypotenuse hence the value will never be more than one. This applies for the cosine function too because $\cos \theta =\dfrac{B}{H}$.