BAM! Mr. Tarrou Well we just got done learning law of sine and law of cosine
we just got done learning that there's more than just one formula for finding
the area of a triangle and we've been working with finding missing parts of
non right triangles we're gonna bring some of those skills together in this
interesting question it's in two parts you'll find did the timestamps in the
description to jump ahead to say Part B if you so desire you'll also find a link
to a website where I have my lessons organized by subject and chapter with H
within each other those subjects instead of scrolling through playlists that are
tens if not hundreds of videos long okay the length of two sides in a triangle
are five centimeters and two centimeters we're going to let theta be the angle
between the two given sides the triangle has an area of two square root of six
centimeters squared we're going to show that the ratio for the sine of theta
this is a non calculator question so we're not going to actually find the
measure of theta but we're going to show that that ratio the sine of theta is
equal to 2 square root of 6 over 5 and then in Part B we're going to find the
two possible values for the third length of the triangle well our information
you'll see actually let's go ahead and just start with that it's always a good
idea to visualize what's actually going on we have a triangle whose two sides
are two and five we have an angle theta be between or included within those two
sides so theta is right here we have this triangle and the area of that
triangle is 2 square root of 6 now we don't have a fixed theta per se we have
a ratio for theta the sine of theta we're going to show that that's equal to
2 square root of 6 over 5 but that might possibly take on some multiple values
opening the door for a couple of different triangles that can give us the
same area of to square root of six so how do we
attack this problem well let's see here whenever you're given a piece of
information that has a formula attached to it good chance you're going to start
with that formula like if I said the distance between two points is 10 you're
going to start off by setting up the distance formula so the area here is
equal to 2 square root of 6 well get my fancy picture-in-picture voila we have
three choices for finding the area of a triangle 1/2 base times height we have
area is equal to 1/2 two sides and the sine of an included angle that's kind of
a giveaway right there right we're trying to work with something of the
sine of theta and then we have Heron's formula where we find s which is half
the perimeter and then work through the area is equal to the square root of s
times s minus a times s minus B times s minus C three choices which one is going
to be helpful which one works with angles and sine functions it's the
second one ok so we are going to say that 2 square root of 6 is equal to 1/2
of the two sides so 2 and 5 times the sine of theta it's a non calculator
question and I'm trying to make this question my own so it works out you know
fairly cleanly very clearly actually 1/2 of 2 is equal to 1 we divide both sides
by 5 and we have that 2 square root of 6 over 5 is indeed the ratio that we have
for the sine of theta so the first question starts off with recognizing
that you're given information about area go pull that formula out that you know
is most related to the information that is given so getting on to the more
interesting part of this question Part B find the two possible values for the
length of the third side so we can have you know sort of a triangle that looks
like this or maybe there's some other angle
measure that gives us the same area maybe we can let this length of five and
open up that angle so that you know you would think that maybe as you open this
angle up that the triangles going to become larger and therefore hold more
Ariel in that area and that's true but eventually as that angle continues to
slide back open that area is going to start to shrink back down to really
effectively zero once that angle becomes a hundred eighty degrees or PI radians
you would have on the verge of basically having a line segment where there's just
barely a little sliver of an opening so with two and five there's you know
another angle measure not that we're going to find the actual angle measure
that will give you that same area of two square root of six how we're going to
figure that out well again you have a question here that is asking for a
missing part a third unknown side of a potentially non right triangle so what
did we learn that helped us find missing parts of basically oblique or non right
triangles leaks really the appropriate word but now I'm right non right
triangle that's either law of sine in law or law of cosine law of sine or law
of cosine simple sohcahtoa sine such as sine is equal to sine of theta is equal
to opposite over adjacent Sonne good for a right triangle well look at what we
have here we have a triangle with a known side of two a known side of five
we're working with some kind of angle we know something about it the sine of that
angle is two square root of 6 over five and we're looking for these missing side
opposite that angle well that's not you can't solve for that third side but
using law of sine you're going to be using law of cosine so the law of cosine
says that we have some side square is equal to the other two sides squared
and added together - two times those other two sides basically times the
cosine of the angle that's opposite that unknown side that you're trying to solve
or maybe you know the opposite side and you're looking for the angle measure the
side and let's call it a opposite this angle is going to be equal to the other
two sides squared netted together so two squared plus five squared minus two
times two times five times the cosine of angle a or theta okay so we have almost
everything we need to find some information about that third unknown
side but we need to know what the cosine of theta is equal to we know what the
sine of theta is how we're going to use that to find the possible ratios for the
cosine of theta what identity allows you to link those two together you can draw
a diagram which actually I prefer to do for many many many of my questions in
trigonometry but you also have that fat green those Aggron identities sine
squared theta plus cosine squared theta equals one one plus tangent squared
theta is equal to secant squared theta and 1 plus cosecant our cotangent
squared theta is equal to cosecant of theta so we are going to be using the
fact that we have our Pythagorean identity saying that sine squared theta
plus cosine squared theta is equal to 1 sine of theta is 2 square root of 6 over
5 squared now we have plus cosine squared theta equals 1
two squared is four square root of six squared is 6 and four times six is equal
to 24 over 5 squared which is 25 plus the cosine squared theta equals one
subtracting that 24 over 25 over to the other side we have an opt erases to have
more room but we have cosine squared theta it is equal to let's find common
denominators the 1 is going to be 25 over 25 minus 24 over 25 cosine squared
theta then is equal to 1 over 25 and
square rooting both sides of this equation tells us that the cosine of
theta is going to take on the ratio of either 1 over 25 or negative 1 over oksu
be positive or negative 1 over 5 not 1 over 25 does it make sense that we can
have a ratio for the cosine function that is both positive and negative well
if we visualize a X love to kind of talk this a little bit but move this around
but if we have an angle that is acute then we have a sine ratio which is in
terms of angles in standard position Y over R okay in Quadrant 1 your Y values
are positive and your X values cosine is x over R would be positive if you open
that angle up to be obtuse then you can still have a sine ratio which is
positive with sine being you know with the rates equal to sine of theta is
equal to Y over R when you're talking about angles in standard position in
Quadrant two your sine ratios can still be positive where the cosine which is x
over R is negative so yes it makes perfect sense that while the sine of
theta is 2 square root of 6 over 5 meaning that theta can be in either
quadrant 1 or 2 the cosine of that at that angle measure may potentially be
positive or negative so we're going to work with these two
ratios so all we have to do now is a little bit of careful arithmetic what
about four let's color code this how about four cosine of theta being
positive 1/5 well then we have a squared is equal to 4 plus 25 2 squared and 5
squared minus 2 times 2 is 4 times 5 is 20 so minus 20 and then we're going to
let the cosine of theta substitute that cosine of theta value and substitute in
that 1/5 we have a squared is equal to 4 minus 20 divided by 5 is 4 so 4 minus 4
is 0 we have a squared is 25 and therefore we have a is equal to 5 now
this time is I square root both sides of this equation I'm not going to consider
both the positive a negative answer because a here is a length of a triangle
and there's no such thing as a negative length so we'll have to make sure not to
account for that as I look for some orange chalk okay we'll just use red so
now we're going to take into account that the cosine of theta can be negative
1/5 again 2 squared is 4 5 squared is 25 2 times 2 is 4 times 5 is 20 so minus 20
but we're taking out the variable of cosine of theta and plugging in what it
potentially can be equal to which is negative 1/5 well now we have negative
20 divided by negative 5 is positive 4 we have 4 plus 4 which is what 28 and
now or try that again 4 plus 4 which is 8 and 8 plus 25 is 33
thus the other possible length of the third side is a is equal to the square
root of 33 I miss a true Pam go to your homework
you