Mathematics, 07.03.2020 02:43 Olaf2018
In the triangle pictured, let A, B, C be the angles at the three vertices, and let a, b,c be the sides opposite those angles. According to the ``law of sines,'' you always have: b a = sinB sinA Suppose that a and b are pieces of metal which are hinged at C. At first the angle A is π/4 radians=45o and the angle B is π/3 radians = 60o. You then widen A to 46o, without changing the sides a and b. Our goal in this problem is to use the tangent line approximation to estimate the angle B.
Answers: 1
![\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}}](/tpl/images/0537/1969/a3167.png)
![\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}]](/tpl/images/0537/1969/59a3f.png)
![B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]](/tpl/images/0537/1969/c6eaf.png)
with respect to A to get,
, the value of 
![f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]](/tpl/images/0537/1969/efb09.png)
![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}](/tpl/images/0537/1969/0bbb1.png)
from part b![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}](/tpl/images/0537/1969/1faf4.png)
, B is approximately,![B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°](/tpl/images/0537/1969/08a16.png)
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In the triangle pictured, let A, B, C be the angles at the three vertices, and let a, b,c be the sid...
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