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>> This is YourMathGal
Julie Harland.
Please visit my website
at yourmathgal.com where all
of my videos are organized
by topic.
Alright, we're going
to do a problem
where we evaluate the six trig
functions for an angle
and a right triangle.
This is part two.
In part one we learned the
definitions
of the trigonometric functions
of an acute angle A shown
below here where adj stands
for the leg adjacent
to this angle A, opp the leg
of the triangle opposite
to angle A,
and hyp hypotenuse.
So there are these ratios,
right, fractions.
So you either have
to memorize that,
or we had something called
SOCATOA which some people
remember it this way,
sine is opposite
over hypotenuse,
cosine adjacent
over hypotenuse,
and tangent is opposite
over adjacent.
And then we have the
reciprocal ones here below.
Alright, so here's
our problem.
In a right triangle one leg is
6 and the hypotenuse is 9.
Find the values
of all six trig functions
for the angle A adjacent
to the leg of length 6.
Alright, so I don't really
know if 6 is the shorter leg
or the longer leg,
but I'm just going to write
down what I know here.
I've got one leg
so I'm just going
to say leg one I know is 6.
I don't know leg two,
but I do know the hypotenuse
is 9.
So I know by the Pythagorean
theorem --
let's see,
I don't know what this is,
I'll call it X for the moment.
I know that one leg squared
plus the other leg squared
equals the hypotenuse squared.
So this gives me 36 plus X
squared is 81.
You're going to subtract 36
from both sides.
So that gives me X squared
is 45.
So X is the square root of 45.
Now, what's the square of 45?
You could do
that in your calculator
or you could estimate it.
Squared of 49 is 7
so this is a little bit less
than 7.
So I'm just going to keep
in mind this is bigger
than that leg, right?
It's close to 7.
And if we're going
to rationalize it remember
that is the square root
of 9 times 5 so 3 square roots
of 5 is what that would be
if I rationalize it.
So now notice I could draw a
picture now.
So I draw a picture here.
I've got let's say this is 6
and this is a little bit
bigger than 6.
And if this is 6 this is 3
square roots of 5, right,
and the hypotenuse of 9.
Okay? And what it says is my
angle A is adjacent to the leg
of length 6.
So we must have A right here
next to the leg of 6.
So now I know what
that picture looks like.
I can find
out all my six trig functions.
Alright, so here's
our picture.
In case you can't remember I'm
putting the SOCATOA up here.
So what's the sine?
Either you've memorized it's
the opposite over hypotenuse
or maybe you used this nemonic
SOCATOA, SOH sign is opposite
over hypotenuse.
So I have to find A,
look at the leg opposite
which is over here
at 3 square roots of 5.
That goes in the numerator,
right?
Opposite goes
in the numerator.
So 3 square roots of 5
over the hypotenuse
which is 9.
And that could be simplified.
You can cancel the 3 and the 9
so that's just square root
of 5 over 3.
Okay, cosine.
That's adjacent
over hypotenuse.
So here's the side adjacent,
6, and it's
over the hypotenuse
which is 9,
and that also happens
to reduce, right?
You could divide the top
and bottom by 3
so that's two thirds.
So that's the cosine
of A. Alright,
and what about the tangent?
So the tangent is the opposite
over the adjacent.
So the leg opposite is
over here at 3 square roots
of 5.
And the side adjacent is 6.
And, again,
you can reduce the 3 and 6
so you'll have a 2
in the denominator,
square root of 5 over 2.
So those are the values
for the sine, the cosine
and the tangent.
Now, let's do the
reciprocal values.
You don't even have to look
up here again.
You could just take the
reciprocal
of what you see here.
So the cosecant
of A you just do the
reciprocal of 3
over square roots of 5,
and then you could rationalize
that by multiplying square
root of 5 over square root
of 5 to get 3 square roots
of 5 over 5.
For the reciprocal
of cosine is the secant
of A. Now,
it's going to be easier.
The reciprocal
of two thirds is three halves.
And for the cotangent
of A you do the reciprocal
so it will be 2
over square roots of 5.
And, again,
you can rationalize the
denominator
to get 2 square roots
of 5 over 5.
Now, you could have looked
back up at the picture
to get the secant.
Like if you look up here
at the picture you want the
secant of A.
If you remember the secant is
the hypotenuse
over the adjacent side you
would write 9,6, for instance,
and it would still reduce
to 3 halves.
But since I already did the
cosine, after I already
simplified the cosine,
I go ahead
and just do the reciprocal.
It's a little faster.
You should get the
same answer.
Alright, so there's an example
of you're given some
information about a triangle,
and then you're able
to find the various
trig values.
This is YourMathGal
Julie Harland.
Please visit my website
at yourmathgal.com where all
of my videos are organized
by topic.