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

6 Answers
Apr 14, 2015

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#.

Jul 29, 2017

Answer:

#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°#

Jul 29, 2017

Answer:

#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#

Aug 4, 2018

Answer:

#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!

Aug 4, 2018

Answer:

#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#.

Aug 4, 2018

Answer:

#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.

https://www.slideshare.net/GMATPrepNow_free/gmat-geometry-everything-you-need-to-know

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#