Chapter 7.2, Problem 63E

To determine

To show: that $\alpha +\beta =\gamma$ , and to find $\tan \gamma$ .

Answer to Problem 63E

  $\tan \gamma =\frac{17}{6}$

Explanation of Solution

Given information:

Given figure,

  

Calculation:

Consider the following figure:

  

To prove that $\alpha +\beta =\gamma$ , it is required to find the angles $\angle DBQ$ and $\angle DQB$ in the triangle $BQD$ .

From the figure, it is observed that the angles $\angle DBQ$ and $\angle ABC$ are made by intersecting the two lines $AQ$ and $CD$ .

Thus, $\angle DBQ$ and $\angle ABC$ are vertical angles and they are equal.

That is, $\angle DBQ=\angle ABC$

From the figure, observe the triangle $ABC$ .

Then,

  $\begin{array}{c}\angle CAB+\angle ABC+\angle BCA=\pi \\ \alpha +\angle ABC+\frac{\pi }{2}=\pi \\ \angle ABC=\frac{\pi }{2}-\alpha \end{array}$

Since, $\angle DBQ=\angle ABC$ then $\angle DBQ=\frac{\pi }{2}-\alpha$ .

From the figure, it is observed that the angles $\angle DQB$ and $\angle PQR$ are made by intersecting the two lines $BR$ and $PD$ .

Thus, $\angle DQB$ and $\angle PQR$ are vertical angles and they are equal.

That is, $\angle DQB=\angle PQR$

From the figure, observe the triangle $PQR$ .

Then,

  $\begin{array}{c}\angle PQR+\angle QRP+\angle RPQ=\pi \\ \beta +\angle PQR+\frac{\pi }{2}=\pi \\ \angle PQR=\frac{\pi }{2}-\beta \end{array}$

Since, $\angle DQB=\angle PQR$ then $\angle DQB=\frac{\pi }{2}-\beta$ .

From the figure, in the triangle $BQD$ , $\angle DBQ+\angle DQB+\angle BDQ=\pi$

Substitute the values $\angle DBQ=\frac{\pi }{2}-\alpha ,\angle DQB=\frac{\pi }{2}-\beta$ and $\angle BDQ=\gamma$ , then

  $\begin{array}{c}\frac{\pi }{2}-\alpha +\frac{\pi }{2}-\beta +\gamma =\pi \\ \pi -\alpha -\beta +\gamma =\pi \\ \alpha +\beta =\lambda \end{array}$

Therefore, it is proved that $\alpha +\beta =\lambda$ .

Since, $\alpha +\beta =\lambda$ , then $\tan \gamma =\tan \left(\alpha +\beta \right)$ .

Use the formula, $\tan \left(s+t\right)=\frac{\tan s+\tan t}{1-\tan s\tan t}$ , then

  $\tan \gamma =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ ………..(1)

Observe the figure, from the triangle $ABC$ , get that $\tan \alpha =\frac{4}{6}$ and from the triangle $PQR$ , get that $\tan \beta =\frac{3}{4}$ .

Substitute these values in equation (1), then

  $\begin{array}{c}\tan \gamma =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ =\frac{\frac{4}{6}+\frac{3}{4}}{1-\frac{4}{6}\times \frac{3}{4}}\\ =\frac{\frac{16+18}{24}}{\frac{24-12}{24}}\\ =\frac{17}{6}\end{array}$

Therefore, $\tan \gamma =\frac{17}{6}$ .