Question

Find the acute angles A and B if, $\sin \left( A+2B \right)=\dfrac{\sqrt{3}}{2}$ and $\cos \left( A+4B \right)=0,A>B$

Answer 1

Hint:First we will find for what angle of sin we get $\dfrac{\sqrt{3}}{2}$ and for what angle of cos we get 0, and then we will use the formula for general solution of sin and cos and find the angles such that A and B are acute angles. Then we will solve those two equations in two variables and find the value of A and B.

Complete step-by-step answer:
Let’s start our solution,
First we will solve for $\sin \left( A+2B \right)=\dfrac{\sqrt{3}}{2}$,
We know that $\sin 60=\dfrac{\sqrt{3}}{2}$ ,
Hence comparing it with $\sin \left( A+2B \right)=\dfrac{\sqrt{3}}{2}$ we get,
$\sin \left( A+2B \right)=\sin 60$
We know that $60{}^\circ =\dfrac{\pi }{3}$ .
Now we will use the formula for general solution of sin,
Now, if we have $\sin \theta =\sin \alpha$ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$
Now using the above formula for $\sin \left( A+2B \right)=\sin 60$ we get,
$A+2B=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{3} \right)$
Here n = integer.
Now as it is given that A and B must be acute we will take n = 0.
Hence considering the angle in degrees, we get,
$A+2B=60........(1)$
Now we will solve for $\cos \left( A+4B \right)=0$,
We know that $\cos 90=0$ ,
Hence comparing it with $\cos \left( A+4B \right)=0$ we get,
$\cos \left( A+4B \right)=\cos 90$
We know that $90{}^\circ =\dfrac{\pi }{2}$ .
Now we will use the formula for general solution of cos,
Now, if we have $\cos \theta =\cos \alpha$ then the general solution is:
$\theta =2n\pi \pm \alpha$
Now using the above formula for $\cos \left( A+4B \right)=\cos 90$ we get,
$A+4B=2n\pi \pm \dfrac{\pi }{2}$
Here n = integer.
Now as it is given that A and B must be acute we will take n = 0.
Hence considering the angle in degrees, we get,
$A+4B=90........(2)$
Now (2) – (1) we get,
$\begin{aligned} & A+4B-A-2B=90-60 \\\ & 2B=30 \\\ & B=15 \\\ \end{aligned}$
Putting the value of B in equation (1) we get,
$\begin{aligned} & A+2\times 15=60 \\\ & A=30 \\\ \end{aligned}$
Hence, we have found the value of A and B, given A>B.

Note: Another method to solve this question is to expand the given expression using the formula, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$, and then one can try to find out the value of A and B from the two equations they get, and that answer will be the same that we have got.Also students should remember important trigonometric standard angles and formulas for solving these type of questions.