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 adjacent side. However, we are missing the opposite side.

To find the opposite side length we will substitute

b = 14

and

c = 26

into the Pythagorean Theorem and solve for

a .

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

SubstituteII

$b = 14$ , $c = 26$

a^{2} + 14^{2} = 26^{2}

▼

Solve for

a

CalcPow

Calculate power

a^{2} + 196 = 676

SubEqn

$LHS - 196 = RHS - 196$

a^{2} = 480

SqrtEqn

$LHS = RHS$

a = 480

SplitIntoFactors

Split into factors

a = 16 \cdot 30

SqrtProd

$a \cdot b = a \cdot b$

a = 16 \cdot 30

CalcRoot

Calculate root

a = 430

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
$sin θ = \frac{opp}{hyp}$	$sin θ = \frac{4 3 0}{2 6}$	$sin θ = \frac{2 3 0}{1 3}$
$cos θ = \frac{adj}{hyp}$	$cos θ = \frac{1 4}{2 6}$	$cos θ = \frac{7}{1 3}$
$tan θ = \frac{opp}{adj}$	$tan θ = \frac{4 3 0}{1 4}$	$tan θ = \frac{2 3 0}{7}$
$csc θ = \frac{hyp}{opp}$	$csc θ = \frac{2 6}{4 3 0}$	$csc θ = \frac{1 3 3 0}{6 0}$
$sec θ = \frac{hyp}{adj}$	$sec θ = \frac{2 6}{1 4}$	$sec θ = \frac{1 3}{7}$
$cot θ = \frac{adj}{opp}$	$cot θ = \frac{1 4}{4 3 0}$	$cot θ = \frac{7 3 0}{6 0}$

Exercise

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