At max height $H = \dfrac{u^2}{4g}$; horizontal distance $= \dfrac{R}{2} = \dfrac{u^2}{2g}$

Elevation angle: $\tan\phi = \dfrac{H}{R/2} = \dfrac{u^2/(4g)}{u^2/(2g)} = \dfrac{1}{2}$

$\phi = \mathbf{\tan^{-1}\!\left(\dfrac{1}{2}\right)}$

Only errors affecting individual supplier accounts change the purchases ledger total.
1. Returns outwards omitted: $57,400 - $350 = $57,050.
The others (Discount allowed, payments undercast in cash book, journal overcast) affect the Control Account or other ledgers, but not the individual ledger balances until corrected there.

We know that $c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}$, where $c$ is the speed of light with dimension $[LT^{-1}]$.

Therefore: $\sqrt{\mu_0\varepsilon_0} = \dfrac{1}{c}$

$[\sqrt{\mu_0\varepsilon_0}] = [c^{-1}] = [L^{-1}T]$

So $(\mu_0\varepsilon_0)^{1/2}$ has dimensions $[L^{-1}T]$.

Total cost $= 19{,}00{,}000 + 1{,}00{,}000 = Rs.\ 20{,}00{,}000$

Depreciable amount $= 20{,}00{,}000 - 40{,}000 = Rs.\ 19{,}60{,}000$

Annual depreciation (SLM) $= \dfrac{19{,}60{,}000}{5} = Rs.\ 3{,}92{,}000$

Under the Straight Line Method, depreciation is equal every year. So the 4th year's depreciation is also Rs. 3,92,000. There is no change between years in SLM.

For the parabola $y^2=4\sqrt{5}x$: tangent in slope form is $y=mx+\frac{\sqrt{5}}{m}$.

For the circle $x^2+y^2=5/2$ (radius $\sqrt{5/2}$): distance from origin to tangent $= \frac{|\sqrt{5}/m|}{\sqrt{1+m^2}}=\sqrt{5/2}$

$\frac{5/m^2}{1+m^2}=\frac{5}{2}\Rightarrow\frac{1}{m^2(1+m^2)}=\frac{1}{2}\Rightarrow m^2+m^4=2\Rightarrow m^4+m^2-2=0$

Hmm: $m^4+m^2-2=(m^2+2)(m^2-1)=0$, so $m=\pm1$. Statement II gives $m^4-3m^2+2=0=(m^2-1)(m^2-2)=0$, so $m=\pm1,\pm\sqrt{2}$ — this is different. But $m=\pm1$ works for both. At $m=1$: tangent is $y=x+\sqrt{5}$ ✓. Statement I is true; Statement II is false (wrong equation).

Let $\cos^{-1}x=A$ and $\cos^{-1}(y/2)=B$, so $A-B=\alpha$.

Then $\cos A=x,\ \cos B=y/2,\ \sin A=\sqrt{1-x^2},\ \sin B=\sqrt{1-y^2/4}$.

Using $\cos\alpha=\cos(A-B)=\cos A\cos B+\sin A\sin B=\dfrac{xy}{2}+\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}$

$4x^2-4xy\cos\alpha+y^2 = (2x-y\cos\alpha)^2+y^2(1-\cos^2\alpha)$... expanding directly: $= 4x^2+y^2-4xy\cos\alpha$.

After substitution and simplification using the expression for $\cos\alpha$: $= 4(1-\cos^2\alpha) = \mathbf{4\sin^2\alpha}$.