Chapter 4, Problem 42RE

Solution

To solve: the given right triangle.

  $\alpha ={44.4153}^{\circ },\beta ={45.5847}^{\circ },a=2\sqrt{6}$ .

Given:

  

  $b=5,c=7$ .

Concept used:

Pythagorean Theorem:

In any right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the hypotenuse.

The two acute angles of a right triangle are complementary.

In any right triangle,

  $\sin \theta =\frac{\text{opp}}{\text{hyp}},\cos \theta =\frac{\text{adj}}{\text{hyp}},\tan \theta =\frac{\text{opp}}{\text{adj}},\csc \theta =\frac{\text{hyp}}{\text{opp}},\sec \theta =\frac{\text{hyp}}{\text{adj}},\cot \theta =\frac{\text{adj}}{\text{opp}}$ .

Calculation:

In the triangle ABC,

  $\begin{array}{c}\cos \alpha =\frac{b}{c}\\ \cos \alpha =\frac{5}{7}\\ \alpha ={\cos }^{-1}\left(\frac{5}{7}\right)\\ \alpha =\text{44}{\text{.4153}}^{\circ }\end{array}$

Here the angles $\alpha$ and $\beta$ are the acute angles of the right triangle ABC.

So,

  $\begin{array}{c}\alpha +\beta ={90}^{\circ }\\ {44.4153}^{\circ }+\beta ={90}^{\circ }\\ \beta ={45.5847}^{\circ }\end{array}$

Now by using the Pythagorean Theorem in the triangle ABC, it follows that

  $\begin{array}{c}{a}^2+b^2=c^2\\ a^2+5^2=7^2\\ a^2=24\\ a=2\sqrt{6}\end{array}$