Exercise 8 Page 466

We want to evaluate six trigonometric functions of the angle $θ .$ In a right triangle the hypotenuse is the side that is opposite the right angle. Therefore, in the given right triangle we have the hypotenuse and the opposite side. However, we are missing the adjacent side.

To find the adjacent side length we will substitute

b = 3

and

c = 9

into the Pythagorean Theorem and solve for

a .

a^{2} + b^{2} = c^{2}

SubstituteII

$b = 3$ , $c = 9$

a^{2} + 3^{2} = 9^{2}

▼

Solve for

a

CalcPow

Calculate power

a^{2} + 9 = 81

SubEqn

$LHS - 9 = RHS - 9$

a^{2} = 72

SqrtEqn

$LHS = RHS$

a = 72

SplitIntoFactors

Split into factors

a = 36 \cdot 2

SqrtProd

$a \cdot b = a \cdot b$

a = 36 \cdot 2

CalcRoot

Calculate root

a = 62

Note that when solving the equation we only kept the principal root. This is because

a

is the side length of a triangle and must be positive.

Let's find the values of the six trigonometric functions for angle $θ .$ Remember to rationalize denominators if needed.

Function	Substitute	Simplify
$sin θ = \frac{opp}{hyp}$	$sin θ = \frac{3}{9}$	$sin θ = \frac{1}{3}$
$cos θ = \frac{adj}{hyp}$	$cos θ = \frac{6 2}{9}$	$cos θ = \frac{2 2}{3}$
$tan θ = \frac{opp}{adj}$	$tan θ = \frac{3}{6 2}$	$tan θ = \frac{2}{4}$
$csc θ = \frac{hyp}{opp}$	$csc θ = \frac{9}{3}$	$csc θ = 3$
$sec θ = \frac{hyp}{adj}$	$sec θ = \frac{9}{6 2}$	$sec θ = \frac{3 2}{4}$
$cot θ = \frac{adj}{opp}$	$cot θ = \frac{6 2}{3}$	$cot θ = 22$