## 15 Degree Triangle

#### Problem

In the diagram below, $AB$ represents the diameter, $C$ lies on the circumference of the circle, and you are given that (Area of Circle) / (Area of Triangle) = $2\pi$.

Prove that the two smaller angles in the triangle are exactly $15^o$ and $75^o$ respectively.

#### Solution

We shall consider three quite different solutions to this problem.

**Method 1**

As triangle $ABC$ is in a semi-circle, angle $ACB$ is a right angle.

Let angle $CBA = \theta, AB = 2r, AC = 2r sin(\theta),$ and $BC = 2r cos(\theta)$.

$\therefore A\Delta = \frac{1}{2}4r^2 sin(\theta) cos(\theta) = 2r^2 sin(\theta) cos(\theta),$ and using the double angle identity: $sin(2\theta) = 2 sin(\theta) cos(\theta),$ we get, $A\Delta = r^2 sin(2\theta)$.

So, (Area of Circle) / (Area of Triangle) = $\dfrac{\pi r^2}{r^2 sin(2\theta)} = \dfrac{\pi}{sin(2\theta)} = 2\pi$

$\therefore sin(2\theta) = \frac{1}{2} \Rightarrow 2\theta = 30^o \Rightarrow \theta = 15^o$

Hence the complementary angle must be 75 degrees.

**Method 2**

In the diagram, $CD$ is perpendicular to $AB$. Let the diameter, $AB = 2r, OC = r, OD = x,$ and $CD = y$.

$\therefore A\Delta = \frac{1}{2}2ry = ry$

So, (Area of Circle) / (Area of Triangle) $= \dfrac{\pi r^2}{ry} = \dfrac{\pi r}{y} = 2\pi \Rightarrow r = 2y$

Using the Pythagorean Theorem,

$x^2 + y^2 = r^2,$ so $x^2 + y^2 = 4y^2 \Rightarrow x^2 = 3y^2 \Rightarrow x = \sqrt{3}y$

Let angle $DOC = \theta$. Therefore $tan(\theta) = \dfrac{y}{x} = \dfrac{y}{\sqrt{3}y} = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = 30^o$.

By using the result that the angle at the centre of a circle ($AOC$) is twice the angle at the circumference we deduce that angle $ABC = 15^o$, and it follows that the complementary angle must be 75 degrees.

**Method 3**

In triangle $ABC$, let $BC = a, AC = b,$ and $AB = c$.

Because the angle in a semi-circle is a right angle, $A\Delta = \frac{1}{2}ab,$ and using the Pythagorean theorem, $a^2 + b^2 = c^2$.

As radius, $r = \frac{1}{2}c \Rightarrow r^2 = \frac{1}{4}c^2 = \frac{1}{4}\left(a^2+b^2\right)$.

So, (Area of Circle) / (Area of Triangle) $= \dfrac{ \pi\frac{1}{4}\left(a^2+b^2\right) }{ \frac{1}{2}ab } = 2\pi$

$$\begin{eqnarray}\therefore \dfrac{a^2+b^2}{2ab} & = & 2\\\therefore a^2+b^2 & = & 4ab\\\therefore \dfrac{a}{b} + \dfrac{b}{a} & = &4\end{eqnarray}$$But in triangle $ABC$, $tan(A) = \dfrac{a}{b}$ and $tan(B) = \dfrac{b}{a},$ so $tan(A) + tan(B) = 4$.

However, $tan(B) = tan(90 - A) = \dfrac{1}{tan(A)}$. Therefore, $tan(A) + \dfrac{1}{tan(A)} = 4$.

By letting $t = tan(A),$ we get $t + \dfrac{1}{t} = 4$. This leads to the quadratic, $t^2 - 4t + 1 = 0,$ which has roots $t = tan(A) = 2 \pm \sqrt{3}$.

We will now show that the root corresponding with $tan(A) = 2 - \sqrt{3}$ is exactly 15 degrees.

Let $T = tan(15)$ and by using the trigonometric identity, $tan(2x) = \dfrac{2tan(x)}{1 - tan^(x)},$ we get $tan(30) = \dfrac{1}{\sqrt{3}} = \dfrac{2T}{1-T^2}$. This leads to the quadratic, $T^2 + 2\sqrt{3}T - 1 = 0,$ which has two roots, $T = -\sqrt{3} \pm 2$. However, as $tan(15) \gt 0,$ we take the positive root and deduce that $tan(15) = 2 - \sqrt{3}$.

Hence the complementary angle must be 75 degrees.