Geometry and Trigonometry

The length of the hypotenuse is $6$, and the lengths of the legs are $x$ and $a$. According to the Pythagorean Theorem:

$$x^2 + a^2 = 6^2$$ $$x^2 + a^2 = 36$$

Subtract $a^2$ from both sides.

$$x^2 = 36 - a^2$$ $$x = \pm\sqrt{36 - a^2}$$

As a length, $x$ must be positive. Hence,

$$x = \sqrt{36 - a^2}$$

We know that $\pi$ radians is equal to $180°$. Hence,

$$\frac{2\pi}{3} = \frac{2 \cdot 180°}{3} = 120°$$

Let’s assume the Earth is approximately a sphere. Thus, using the formula for the volume of a sphere we get

$$2.6 \cdot 10^{11} = \frac{4}{3}\pi r^3$$

where $r$ is the radius. Let’s solve for $r$.

Multiply both sides by $3$.

$$3 \cdot 2.6 \cdot 10^{11} = 4\pi r^3$$

Divide both sides by $4\pi$.

$$\frac{3 \cdot 2.6 \cdot 10^{11}}{4\pi} = r^3$$ $$6.20704\ldots \cdot 10^{10} = r^3$$ $$r = \sqrt[3]{6.20704\ldots \cdot 10^{10}}$$ $$r = 3959.389\ldots \approx 3960$$

So the radius is approximately $3960$ miles.

Let’s look at the triangles $ABC$ and $ADE$ in the picture. They are both right triangles (angles $ADE$ and $ABC$ are right angles) and they both share the same angle $A$. Hence, by the angle-angle rule, they are similar triangles.

The leg $DE$ in triangle $ADE$ corresponds to the leg $BC$ in triangle $ABC$, and the leg $AD$ in triangle $ADE$ corresponds to the leg $AB$ in triangle $ABC$. Their lengths, in feet, are:

$$|DE| = 8 \qquad |BC| = 12 \qquad |AD| = 15 - x \qquad |AB| = 15$$

Hence, we can form a proportion and solve for $x$.

$$\frac{15 - x}{8} = \frac{15}{12}$$

Cross multiply.

$$(15 - x) \cdot 12 = 15 \cdot 8$$ $$15 \cdot 12 - x \cdot 12 = 15 \cdot 8$$ $$180 - 12x = 120$$

Subtract $120$ from both sides.

$$60 - 12x = 0$$

Add $12x$ to both sides.

$$60 = 12x$$

Divide both sides by $12$.

$$\frac{60}{12} = x$$ $$x = 5$$