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## Trigonometry

### Unit 1: Lesson 3

Solving for a side in a right triangle using the trigonometric ratios# Solving for a side in right triangles with trigonometry

CCSS.Math:

Sal is given a right triangle with an acute angle of 65° and a leg of 5 units, and he uses trigonometry to find the two missing sides. Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

We're asked to solve the
right triangle shown below. Give the lengths to
the nearest tenth. So when they say solve
the right triangle, we can assume that
they're saying, hey figure out the
lengths of all the sides. So whatever a is equal to,
whatever b is equal to. And also what are all the
angles of the right triangle? They've given two of them. We might have to figure out
this third right over here. So there's multiple
ways to tackle this, but we'll just try to
tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator,
and using a calculator, you can use your
trigonometric functions that we've looked
at a good bit now. So I'll give you a
few seconds to think about how to figure
out what a is. Well, what do we know? We know this angle
y right over here. We know the side
adjacent to angle y. And length a, this
is the side that's the length of the side that
is opposite to angle Y. So what trigonometric
ratio deals with the opposite
and the adjacent? So if we're looking at angle
Y, relative to angle Y, this is the opposite. And this right over
here is the adjacent. Well if we don't remember,
we can go back to SohCahToa. Sine deals with
opposite and hypotenuse. Cosine deals with
adjacent and hypotenuse. Tangent deals with
opposite over adjacent. So we can say that the
tangent of 65 degrees, of that angle of 65 degrees,
is equal to the opposite, the length of the
opposite side, which we know has length a over the
length of the adjacent side, which they gave us in the
diagram, which has length five. And you might say,
how do I figure out a? Well we can use our
calculator to evaluate what the tangent
of 65 degrees are. And then we can solve for a. And actually if we just want to
get the expression explicitly solving for a, we
could just multiply both sides of this
equation times 5. So let's do that. 5 times, times 5. These cancel out,
and we are left with, if we flip the
equal around, we're left with a is equal to 5 times
the tangent of 65 degrees. So now we can get our
calculator out and figure out what this is to
the nearest tenth. That's my handy TI-85 out and
I have 5 times the tangent-- I didn't need to press that
second right over there, just a regular
tangent-- of 65 degrees. And I will get, if I
round to the nearest tenth like they ask me to, I get 10.7. So a is approximately
equal to 10.7. I say approximately
because I rounded it down. This is not the exact number. But a is equal to 10.7. So we now know that this has
length 10.7, approximately. There are several ways that
we can try to tackle b. And I'll let you pick
the way you want to. But then I'll just do it
the way I would like to. So my next question
to you is, what is the length of the side YW? Or what is the value of b? Well there are
several ways to do it. This is the hypotenuse. So we could use
trigonometric functions that deal with adjacent
over hypotenuse or opposite over hypotenuse. Or we could just use
the Pythagorean theorem. We know two sides
of a right triangle. We can come up with
the third side. I will go with using
trigonometric ratios since that's what we've
been working on a good bit. So this length of b, that's
the length of the hypotenuse. So this side WY
is the hypotenuse. And so what trigonometric
ratios-- or we can decide what we want to use. We could use opposite
and hypotenuse. We could use adjacent
and hypotenuse. Since we know that
XY is exactly 5 and we don't have to deal
with this approximation, let's use that side. So what trigonometric
ratios deal with adjacent and hypotenuse? Well we see from
SohCahToa cosine deals with adjacent over hypotenuse. So we could say that
the cosine of 65 degrees is equal to the length of
the adjacent side, which is 5 over the length of
the hypotenuse, which has a length of b. And then we can
try to solve for b. You multiply both
sides times b, you're left with b times cosine of
65 degrees is equal to 5. And then to solve for b,
you could divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing--
we have to figure it out what our calculator,
but this is just going to evaluate
to some number. So we can divide both sides by
that, by cosine of 65 degrees. And we're left with
b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator
to figure out the length of b. Length of b is 5 divided
by cosine of 65 degrees. And I get, if I round to
the nearest tenth, 11.8. So b is approximately equal
to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done
solving this right triangle. And you could have figured
this out using the Pythagorean theorem as well, saying that
5 squared plus 10.7 squared should be equal to b squared. And hopefully you would
get the exact same answer. And the last thing
we have to figure out is the measure of angle
W right over here. So I'll give you a
few seconds to think about what the
measure of angle W is. Well here we just
have to remember that the sum of the angles of a
triangle add up to 180 degrees. So angle w plus 65 degrees,
that's this angle right up here, plus the right angle,
this is a right triangle, they're going to add
up to 180 degrees. So all we need to do
is-- well we can simplify the left-hand side
right over here. 65 plus 90 is 155. So angle W plus 155 degrees
is equal to 180 degrees. And then we get angle W-- if we
subtract 155 from both sides-- angle W is equal to 25 degrees. And we are done solving the
right triangle shown below.