Question

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

Answer 1

In this way:

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

`sinA=sqrt(1-cos^2A)=sqrt(1-25/169)=sqrt((169-25)/169)=`

`=sqrt(144/169)=12/13`,

and

`tanA=sinA/cosA=(12/13)/(5/13)=12/13*13/5=12/5`.

Answer 2

`sin A = 12/13`

`cos A = 5/13`

`tan A = 12/5`

Explanation:

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

`sin theta = "opposite"/"hypotenuse"`

`cos theta = "adjacent"/"hypotenuse"`

`tan theta = "opposite"/"adjacent"`

So, as we are given `cos A = 5/13`, it means that

in this specific right-angled triangle,

the side adjacent to `A = 5 and "the hypotenuse " = 13`

Using Pythagoras' Theorem.

`opp = sqrt(13^2-5^2) = 12`

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

Now we can give the trig ratios as:

`sin A = "opposite"/"hypotenuse" = 12/13`

`cos A = "adjacent"/"hypotenuse" = 5/13`

`tan A = "opposite"/"adjacent" = 12/5`

From these,we can find that

angle `A = 67.4°`

Answer 3

`sin A = +- 12/13`
`tan A = +- 12/5`

Explanation:

`cos A = 5/13`
`sin^2 A = 1 - cos^2 a = 1 - 25/169 = 144/169`
`sin A = +- 12/13`.
There are 2 opposite values of sin A, because, when cos A = `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 A/(cos A) = +- (12/13)(13/5) = +- 12/5`

Answer 4

`sina=12/13` and `tana=12/5`

Explanation:

If we have a right triangle where `cosa=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:

`sina="opposite"/"hypotenuse"`

`cosa="adjacent"/"hypotenuse"`

`tana="opposite"/"adjacent"`

From this, we see that

`sina=12/13` and `tana=12/5`

Hope this helps!

Answer 5

`sin A = 12/13 and tan A = 5/13`, for `A in Q_1` and
`sin A = - 12/13 and tan A = - 5/13`, for `A in Q_4`,
`A = 2kpi +- 1.176 rad, k = 0, +-1, +-2, +-3, ..`

Explanation:

Unless otherwise restricted,

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

`= 2kpi +- 1.176 rad`, nearly, `in Q_1` or `Q_4`,

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

`= ... - 67.38^o, 67.38^o,. ...`

And so,

`sin A = 12/13 and tan A = 5/13`, for `A in Q_1` and

`sin A = - 12/13 and tan A = - 5/13`, for `A in Q_4`.

Answer 6

`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.

`:. 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`