Chapter 5, Problem 5PS

To determine

Prove your conjecture by using the identity for the tangent of the sum of two angles.

Answer to Problem 5PS

We have verified the conjecture.

Explanation of Solution

Given information:

Three squares of side length $s$ are placed side by side (see figure). Make a conjecture about the relationship between the sum $u+v$ and $w$ . Prove your conjecture by using the identity for the tangent of the sum of two angles.

Calculation:

Let us consider that the three adjacent squares have been kept as below.

  

The relation between the sum $u+v$ and $w$ can be determined by applying Lami’s Theorem or the Sine Law of triangle given below.

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

  

Applying Lami’s Theorem to triangle $BCH$ , where angle $x$ is angle between lines $BH$ and $CH$ .

  $\begin{array}{l}\frac{\sin v}{BH}=\frac{\sin \left(\pi -w\right)}{CH}=\frac{\sin x}{BC}\\ \frac{\sin v}{\sqrt{A{B}^2+A{H}^2}}=\frac{\sin w}{\sqrt{A{C}^2+A{H}^2}}=\frac{\sin x}{BC}\\ \frac{\sin v}{\sqrt{s^2+s^2}}=\frac{\sin w}{\sqrt{4s^2+s^2}}=\frac{\sin x}{s}\\ \frac{\sin v}{\sqrt{2}s}=\frac{\sin w}{\sqrt{5}s}=\frac{\sin x}{s}\end{array}$

We obtain a relation between angle $v$ and $x$ as follows from the above expression.

  $\begin{array}{l}\frac{\sin v}{\sqrt{2}s}=\frac{\sin x}{s}\\ \frac{\sin v}{\sqrt{2}}=\sin x\mathrm{.....}\left(1\right)\end{array}$

Similarly we apply Lami’s Theorem to triangle $DCH$ , where angle $y$ is the angle between lines $DH$ and $CH$ .

  $\begin{array}{l}\frac{\sin u}{CH}=\frac{\sin \left(\pi -v\right)}{DH}=\frac{\sin y}{DC}\\ \frac{\sin u}{\sqrt{A{C}^2+A{H}^2}}=\frac{\sin v}{\sqrt{A{D}^2+A{H}^2}}=\frac{\sin y}{DC}\\ \frac{\sin u}{\sqrt{4s^2+s^2}}=\frac{\sin v}{\sqrt{9s^2+s^2}}=\frac{\sin y}{s}\\ \frac{\sin u}{\sqrt{5}s}=\frac{\sin v}{\sqrt{10}s}=\frac{\sin y}{s}\end{array}$

We obtain a relation between angle $v$ and $u$ as follows from the above expression.

  $\begin{array}{l}\frac{\sin v}{\sqrt{10}s}=\frac{\sin u}{\sqrt{5}s}\\ \frac{\sin v\sqrt{5}}{\sqrt{10}}=\sin u\\ \frac{\sin v}{\sqrt{2}}=\sin u\mathrm{......}(2)\\ \end{array}$

By equation (1) and (2), we can say that

  $\sin u=\sin x$

Since all the angles involved are acute, hence the above expression will hold if and only if

  $x=u\mathrm{.....}(3)$

Considering the triangle $BCH$ again, where by exterior angle property,

  $w=x+v$

By equation (3) and the above expression,

  $w=\boxed{u+v}\mathrm{....}(4)$

The above conjecture can be proved by using sum formula for tangent for the right hand side of the equation (4) given as follows:

  $\begin{array}{l}\tan \left(u+v\right)=\frac{\tan u+\tan v}{1-\tan u\tan v}\\ =\frac{\frac{AH}{AD}+\frac{AH}{AC}}{1-\frac{AH}{AD}\frac{AH}{AC}}\\ =\frac{\frac{s}{3s}+\frac{s}{2s}}{1-\frac{s}{3s}\frac{s}{2s}}\\ =\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\frac{1}{2}}\\ \end{array}$

  $\begin{array}{l}=\frac{\frac{5}{6}}{1-\frac{1}{6}}\\ =\frac{\frac{5}{6}}{\frac{5}{6}}\\ =1\\ \end{array}$

Similarly the tangent of the left hand side of the equation (4) can be written as follows:

  $\begin{array}{l}\tan w=\frac{AH}{AB}\\ =\frac{s}{s}\\ =1\end{array}$

Since the angles on both the sides of the equation (4) are acute and since the tangent of $w$ and the sum $u+1$ are equal, the arguments of the tangents on both the sides are equal.

Hence, we have verified the conjecture.