Question

an angle a in standard position has a terminal side which passes through the point (-5,8)

Answer 1

STEP - BY - STEP EXPLANATION

What to find?

• sin(α)

,

• cos(α)

,

• tan(α)

,

• cot(α)

,

• sec(α)

,

• csc(α)

GIVEN:

Step 1

Make a sketch.

Step 2

Determine the hypotenuse side.

Using the Pythagoras theorem;

$hypotenuse^2=adjacent^2+opposite^2$$\begin{gathered} hypotenuse=\sqrt{5^2+8^2} \\ \\ =\sqrt{25+64} \\ \\ =\sqrt{89} \end{gathered}$

Step 3

Determine the identities.

We know that;

opposite =8

adjacent=5

hypotenuse = √89

$\begin{gathered} sin(\alpha)=\frac{opposite}{hypotenuse} \\ \\ =\frac{8}{\sqrt{89}} \\ \\ =\frac{8\sqrt{89}}{89} \end{gathered}$

Since sine is positive in the second quadrant, then we leave the answer as it is.

Hence;

$sin(\alpha)=\frac{8\sqrt{89}}{89}$

$\begin{gathered} cos(\alpha)=\frac{adjacent}{hypotenuse} \\ \\ =\frac{5}{\sqrt{89}} \\ \\ =\frac{5\sqrt{89}}{89} \end{gathered}$

In quadrant II, cos is negative.

Hence;

$cos(\alpha)=-\frac{5\sqrt{89}}{89}$

$\begin{gathered} tan(\alpha)=\frac{opposite}{adjacent} \\ \\ =\frac{8}{5} \end{gathered}$

Tangent is negative in quadrant II.

Hence;

$tan(\alpha)=-\frac{8}{5}$

$\begin{gathered} cot(\alpha)=\frac{1}{tan\alpha} \\ \\ =\frac{1}{-\frac{8}{5}} \\ \\ =-\frac{5}{8} \end{gathered}$

Hence,

$cot(\alpha)=-\frac{5}{8}$

$\begin{gathered} sec(\alpha)=\frac{1}{cos(\alpha)} \\ \\ =\frac{1}{-\frac{5}{\sqrt{89}}} \\ \\ =-\frac{\sqrt{89}}{5} \end{gathered}$

Hence;

$sec(\alpha)=-\frac{\sqrt{89}}{5}$

$\begin{gathered} csc(\alpha)=\frac{1}{sin(\alpha)} \\ \\ =\frac{1}{\frac{8}{\sqrt{89}}} \\ \\ =\frac{\sqrt{89}}{8} \end{gathered}$

Hence;

$csc(\alpha)=\frac{\sqrt{89}}{8}$

ANSWER

• sin(α) = 8√89 /89

• cos(α) = -5√89 /89

• tan(α)= - 8/5

• cot(α)= -5/8

• sec(α) = - √89 / 5

• cosec(α) =