Chapter 7.3, Problem 48E

a.

To determine

To select:the values for $A,B\text{ and }C$ and verify that the conclusion is true for your specific values.

Answer to Problem 48E

  $\begin{array}{c}A=120°\\ B=30°\\ C=30°\end{array}$

Explanation of Solution

Given information:

  $\tan A+\tan B+\tan C=\tan A\tan B\tan C$

The angles $A,B\text{ and }C$ are the angles of a non-right triangle.

Calculation:

The sum of interior angles of a triangle is 180 $°$ , so let $A=120°,B=30°\text{ and }C=30°$ . Substitute the values in the given equationand both the sides of equations as follows:

  $\begin{array}{c}\tan A+\tan B+\tan C=\tan A\tan B\tan C\\ \tan 120°+\tan 30°+\tan 30°=\tan 120°\tan 30°\tan 30°\\ -\sqrt{3}+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}=\left(-\sqrt{2}\right)\left(\frac{\sqrt{3}}{3}\right)\left(\frac{\sqrt{3}}{3}\right)\\ -\frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{3}\\ LHS=RHS\end{array}$

Since the equation is verified. Thus, the value of angle $A,B\text{ and }C$ is 120 $°$ , 30 $°$ and 30 $°$ respectively.

b.

To determine

To prove:the theorem.

Answer to Problem 48E

See explanation

Explanation of Solution

Given information:

  $\tan A+\tan B+\tan C=\tan A\tan B\tan C$

The angles $A,B\text{ and }C$ are the angles of a non-right triangle.

  $A=120°,B=30°\text{ and }C=30°$

Calculation:

The sum of interior angles of a triangle is 180 $°$ . The theorem can be proved as follows:

  $\begin{array}{c}A+B+C=180°\\ A+B=180°-C\left[\text{Subtract }C\text{ from each side}\right]\\ \tan \left(A+B\right)=\tan \left(180°-C\right)\left[\text{Take tangent on each side}\right]\\ \frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\tan 180°-\tan C}{1+\tan 180°\tan C}\left[\begin{array}{l}\text{Use sum and different }\\ \text{identity of tangent}\end{array}\right]\\ \frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{0-\tan C}{1+\left(0\right)\tan C}\left[\text{Simplify}\right]\\ \frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C\\ \tan A+\tan B=\left(1-\tan A\tan B\right)\left(-\tan C\right)\left[\begin{array}{l}\text{Multiply }\left(1-\tan A\tan B\right)\\ \text{ on each side}\end{array}\right]\\ \tan A+\tan B=-\tan C+\tan A\tan B\tan C\\ \tan A+\tan B+\tan C=\tan A\tan B\tan C\left[\text{Add }\tan C\text{ on each side}\right]\end{array}$ .

Hence, the theorem is proved.