Chapter 9.1, Problem 6GP

Solution

To solve: the triangle $△ABC$

  $B=58°,a=6.25,c=11.79$

Given information:

The diagram:

  

  $A=32°,b=10$

Definition Used:

Let $\theta$ be an acute angle of a right triangle. The six trigonometric functions of $\theta$ are defined as follows:

  $\sin \theta =\frac{opposite}{hypotenuse}$ , $\cos \theta =\frac{adjacent}{hypotenuse}$ , $\tan \theta =\frac{opposite}{adjacent}$

  $\cot \theta =\frac{adjacent}{opposite}$ , $\csc \theta =\frac{hypotenuse}{opposite}$ , $\sec \theta =\frac{hypotenuse}{adjacent}$

Theorem Used:

Pythagorean Theorem:

In a right triangle, the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse.

  $Hypotenus{e}^2=opposit{e}^2+adjacen{t}^2$

For the given triangle, this is:

  

  $a^2+b^2=c^2$

Explanation:

The given triangle is a right angled triangle, with right angle at C.

This implies that angles A and B are complimentary angles, that is their sum is $90°$ . So,

  $\begin{array}{c}A+B=90°\\ 32°+B=90°\\ B=90°-32°\\ B=58°\end{array}$

Now,

  $\begin{array}{c}\tan A=\frac{opp}{adj}\\ \tan 32°=\frac{a}{b}\\ \tan 32°=\frac{a}{10}\end{array}$

  $\begin{array}{c}a=10\tan 32°\\ a\approx 6.25\end{array}$

  $\begin{array}{c}\sec A=\frac{hyp}{adj}\\ \sec 32°=\frac{c}{b}\\ \sec 32°=\frac{c}{10}\end{array}$

  $\begin{array}{c}c=10\sec 32°\\ c\approx 11.79\end{array}$