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2 years ago

I 23 12 N L 12 23 O K M 18 18

Mathematics

1 answer:

soldi70 [24.7K]2 years ago

8 0

Answer: $\sqrt{407}$

Step-by-step explanation:

In this triangle, we know that $IJ=46, IK=24, JK=36$ .

So, using the Law of Cosines in triangle IJK,

$36^2 = 46^2 + 24^2 - 2(46)(24) \cos (\angle JIK)\\\\-1396=-2(46)(24) \cos(\angle JIK)\\\\\cos (\angle JIK)=\frac{349}{552}$

Using the Law of Cosines in the triangle LIK,

$(KL)^2 = 23^2 + 24^2 - 2(23)(24) \cos (\angle JIK)\\\\(KL)^2 = 23^2 + 24^2 - 2(23)(24)\left(\frac{349}{552} \right)\\\\(KL)^2 = 407\\\\KL=\sqrt{407}$

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Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

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In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$