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If $\cos^{-1}\!\left(\dfrac{2}{3x}\right)+\cos^{-1}\!\left(\dfrac{3}{4x}\right)=\dfrac{\pi}{2}$ for $x>\dfrac{3}{4}$, find $x$.
$\cos^{-1}A+\cos^{-1}B=\pi/2 \Rightarrow \cos^{-1}A=\pi/2-\cos^{-1}B=\sin^{-1}B$, i.e., $A=\sin(\cos^{-1}B)=\sqrt{1-B^2}$.
$\dfrac{2}{3x}=\sqrt{1-\dfrac{9}{16x^2}}$
Square: $\dfrac{4}{9x^2}=1-\dfrac{9}{16x^2}$
Multiply by $144x^2$: $64=144x^2-81 \Rightarrow 144x^2=145 \Rightarrow x=\dfrac{\sqrt{145}}{12}$
How is any premium on the issue of shares treated in the financial statements of a limited company?
Option A is correct.
Share premium is the excess above nominal value when shares are issued. It is recorded in the Share Premium Account, which is a capital reserve — it cannot be distributed as dividends.
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