Answer:
Step-by-step explanation:
(a)
Consider the following:
[tex]A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°[/tex]
Use sine rule,
[tex]\frac{b}{a}=\frac{\sinB}{\sin A} \\\\=\frac{\sin{\frac{\pi}{3}} }{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}[/tex]
Again consider,
[tex]\frac{b}{a}=\frac{\sin{B}}{\sin{A}} \\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}][/tex]
Thus, the angle B is function of A is, [tex]B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}][/tex]
Now find [tex]\frac{dB}{dA}[/tex]
Differentiate implicitly the function [tex]\sin{B}=\sqrt{\frac{3}{2}}\sin{A}[/tex] with respect to A to get,
[tex]\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}[/tex]
b)
When [tex]A=\frac{\pi}{4},B=\frac{\pi}{3}[/tex], the value of [tex]\frac{dB}{dA}[/tex] is,
[tex]\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}[/tex]
c)
In general, the linear approximation at x= a is,
[tex]f(x)=f'(x).(x-a)+f(a)[/tex]
Here the function [tex]f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}][/tex]
At [tex]A=\frac{\pi}{4}[/tex]
[tex]f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}[/tex]
And,
[tex]f'(A)=\frac{dB}{dA}=\sqrt{3}[/tex] from part b
Therefore, the linear approximation at [tex]A=\frac{\pi}{4}[/tex] is,
[tex]f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}[/tex]
d)
Use part (c), when [tex]A=46°[/tex], B is approximately,
[tex]B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°[/tex]
