# If cos A = 5/13, how do you find sinA and tanA?

Apr 14, 2015

In this way:

(Without further information about the angle $A$, I assume that that angle is in the first quadrant)

$\sin A = \sqrt{1 - {\cos}^{2} A} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{169 - 25}{169}} =$

$= \sqrt{\frac{144}{169}} = \frac{12}{13}$,

and

$\tan A = \sin \frac{A}{\cos} A = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{13} \cdot \frac{13}{5} = \frac{12}{5}$.

Jul 29, 2017

$\sin A = \frac{12}{13}$

$\cos A = \frac{5}{13}$

$\tan A = \frac{12}{5}$

#### Explanation:

The basic trig functions are defined in a right-angled triangle as:

$\sin \theta = \text{opposite"/"hypotenuse}$

$\cos \theta = \text{adjacent"/"hypotenuse}$

$\tan \theta = \text{opposite"/"adjacent}$

So, as we are given $\cos A = \frac{5}{13}$, it means that

in this specific right-angled triangle,

the side adjacent to $A = 5 \mathmr{and} \text{the hypotenuse } = 13$

Using Pythagoras' Theorem.

$o p p = \sqrt{{13}^{2} - {5}^{2}} = 12$

So now the side opposite $A$ is $12$

Now we can give the trig ratios as:

$\sin A = \text{opposite"/"hypotenuse} = \frac{12}{13}$

$\cos A = \text{adjacent"/"hypotenuse} = \frac{5}{13}$

$\tan A = \text{opposite"/"adjacent} = \frac{12}{5}$

From these,we can find that

angle A = 67.4°

Jul 29, 2017

$\sin A = \pm \frac{12}{13}$
$\tan A = \pm \frac{12}{5}$

#### Explanation:

$\cos A = \frac{5}{13}$
${\sin}^{2} A = 1 - {\cos}^{2} a = 1 - \frac{25}{169} = \frac{144}{169}$
$\sin A = \pm \frac{12}{13}$.
There are 2 opposite values of sin A, because, when cos A = $\frac{5}{13}$,
the arc (angle) A could be either in Quadrant 1 or in Quadrant 4.
There are also 2 opposite values for tan A
$\tan A = \sin \frac{A}{\cos A} = \pm \left(\frac{12}{13}\right) \left(\frac{13}{5}\right) = \pm \frac{12}{5}$

Aug 4, 2018

$\sin a = \frac{12}{13}$ and $\tan a = \frac{12}{5}$

#### Explanation:

If we have a right triangle where $\cos a = \frac{5}{13}$, this means that the adjacent side is $5$ and the hypotenuse is $13$.

With the Pythagorean Theorem, we find that the opposite side is $12$. Recall SOH-CAH-TOA:

$\sin a = \text{opposite"/"hypotenuse}$

$\cos a = \text{adjacent"/"hypotenuse}$

$\tan a = \text{opposite"/"adjacent}$

From this, we see that

$\sin a = \frac{12}{13}$ and $\tan a = \frac{12}{5}$

Hope this helps!

Aug 4, 2018

$\sin A = \frac{12}{13} \mathmr{and} \tan A = \frac{5}{13}$, for $A \in {Q}_{1}$ and
$\sin A = - \frac{12}{13} \mathmr{and} \tan A = - \frac{5}{13}$, for $A \in {Q}_{4}$,
$A = 2 k \pi \pm 1.176 r a d , k = 0 , \pm 1 , \pm 2 , \pm 3 , . .$

#### Explanation:

Unless otherwise restricted,

A = 2kpi +- arccos ( 5/13 ) = 2kpi +- 67.38^o

$= 2 k \pi \pm 1.176 r a d$, nearly, $\in {Q}_{1}$ or ${Q}_{4}$,

k = 0, +-1, +-2, +-3, ...3

$= \ldots - {67.38}^{o} , {67.38}^{o} , . \ldots$

And so,

$\sin A = \frac{12}{13} \mathmr{and} \tan A = \frac{5}{13}$, for $A \in {Q}_{1}$ and

$\sin A = - \frac{12}{13} \mathmr{and} \tan A = - \frac{5}{13}$, for $A \in {Q}_{4}$.

Aug 4, 2018

color(chocolate)(cos A = (AC) / (AB) = 5 / 13, sin A = (BC) / (AB) = 12 / 13, tan A = (BC) / (AC) = 12/ 5

#### Explanation:

A Pythagorean triple consists of three positive integers a, b, and c, such that a2 + b2 = c2.

If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k. A primitive Pythagorean triple is one in which a, b and c are co-prime (that is, they have no common divisor larger than 1).

The best known triple is 3-4-5, with 5-12-13 being the next most recognised.

Any triangle composed of sides of lengths that match the Pythagorean triple will be a right triangle.

That means our triangle has a 90 degree angle for angle C.

$\therefore a = 5 , b = 12 , c = 13 , {5}^{2} + {12}^{2} = {13}^{2}$

color(chocolate)(cos A = (AC) / (AB) = 5 / 13, sin A = (BC) / (AB) = 12 / 13, tan A = (BC) / (AC) = 12/ 5