## Solution:

Refer to the figure below.

By ratio and proportion of two similar triangles ΔBCD and ΔACE, we have

$\displaystyle \frac{y}{r-x}=\frac{h}{r}$

$\displaystyle y=\frac{h\left(r-x\right)\:}{r}$

## Solution:

$\displaystyle S=bd^{3\:}$

$\displaystyle S=20d^3$

## Solution:

From the formula of the area of a triangle, $\displaystyle A=\frac{1}{2}\cdot a\cdot b\cdot sin\left(\theta \right)$. Also, we know that an interior angle of an equilateral triangle is 60 degrees, and $\displaystyle sin\:60^{\circ} =\frac{\sqrt{3}}{2}$

$\displaystyle A=\frac{1}{2}\cdot x\cdot x\cdot sin\:60^{\circ}$

$\displaystyle A=\frac{1}{2}\cdot x^2\cdot \frac{\sqrt{3}}{2}$

$\displaystyle A=\frac{\sqrt{3}}{4}x^2$

## Solution:

$\displaystyle Distance=Rate\times Time$

$\displaystyle D=\left(60\:km/hr\right)\cdot t$

$\displaystyle D=60t$

## Solution:

$\displaystyle y=tan\left(x+\pi \right)$

$\displaystyle x+\pi =tan^{-1}y$

$\displaystyle x=tan^{-1}y-\pi$

## Solution:

$\displaystyle y=\frac{x^2+3}{x}$

$\displaystyle xy=x^2+3$

$\displaystyle x^2-xy+3=0$

Solve for x using quadratic formula. We have $\displaystyle a=1,\:b=-y,\:and\:c=3$

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}\:}{2a}$

$\displaystyle x=\frac{\cdot -\left(-y\right)\pm \sqrt{\left(-y\right)^2-4\left(1\right)\left(3\right)}}{2\left(1\right)}$

$\displaystyle x=\frac{y\pm \sqrt{y^2-12}}{2}$