Maths Tutorial: Trigonometry SOH CAH TOA (trigonometric ratios)
Summary
TLDRThe video explains the use of trigonometric ratios—sine, cosine, and tangent—in solving right-angle triangle problems. It walks through how to correctly label triangle sides as hypotenuse, opposite, or adjacent based on the given angle. The mnemonic 'SOHCAHTOA' is used to recall the ratios: sine equals opposite over hypotenuse, cosine equals adjacent over hypotenuse, and tangent equals opposite over adjacent. The video also demonstrates examples where these ratios are applied to calculate unknown sides or angles using a calculator, emphasizing proper calculator settings.
Takeaways
- 📐 Trigonometric ratios (sin, cos, tan) are used to calculate unknown angles or sides in a right-angled triangle when enough information is provided.
- ⚠️ You must always ensure the triangle is a right-angled triangle before applying trigonometric ratios.
- ➕ The hypotenuse is always the longest side and opposite the right angle in a triangle.
- 🔄 Depending on the angle you are working with, the other two sides are labeled as opposite (opposite the angle) and adjacent (next to the angle).
- 🔢 SOHCAHTOA helps remember the trigonometric ratios: sin (opposite/hypotenuse), cos (adjacent/hypotenuse), and tan (opposite/adjacent).
- 🧮 To solve for unknown sides or angles, rearrange the trigonometric ratio formulas and use a calculator in degree mode.
- 📊 Example: Given an angle and hypotenuse, use the sine ratio to find the opposite side.
- 📏 Example: Use the cosine ratio when the adjacent and hypotenuse are involved in the calculation.
- 🔍 The inverse sine, cosine, or tangent function (sin⁻¹, cos⁻¹, tan⁻¹) is used to find the unknown angle when the sides are known.
- 📘 Real-world problem: Trigonometry can be applied to solve practical problems like calculating the angle and length of ropes anchoring a flagpole.
Q & A
What are the trigonometric ratios used for?
-Trigonometric ratios (sine, cosine, and tangent) are used to calculate unknown angles or sides in a right-angle triangle, provided enough information about the triangle is given.
When can the trigonometric ratios not be applied?
-Trigonometric ratios cannot be applied if the triangle is not a right-angle triangle. You must have a right angle to use these ratios.
What is the hypotenuse in a triangle, and how do you identify it?
-The hypotenuse is the longest side of a right-angle triangle and is always opposite the right angle.
How do you label the opposite and adjacent sides in a right-angle triangle?
-The opposite side is the side directly opposite the angle of interest, while the adjacent side is the one next to the angle but not the hypotenuse.
What does the acronym SOHCAHTOA represent?
-SOHCAHTOA is a mnemonic to remember the trigonometric ratios: Sine (S) = Opposite/Hypotenuse, Cosine (C) = Adjacent/Hypotenuse, and Tangent (T) = Opposite/Adjacent.
How do you use the sine ratio to find a missing side of a triangle?
-To find a missing side using the sine ratio, use the formula Sin(θ) = Opposite/Hypotenuse. Rearrange to solve for the unknown side, then plug the known values into the equation.
How do you calculate a missing angle using the inverse trigonometric function?
-To find a missing angle, use the inverse sine, cosine, or tangent function. For example, if Sin(θ) = 3/5, use θ = Sin⁻¹(3/5) to calculate the angle.
What do you need to check before applying trigonometric ratios?
-Before applying trigonometric ratios, always check if the triangle has a right angle. Without a right angle, these ratios cannot be used.
How do you find the angle between a rope and the ground in a real-world scenario?
-To find the angle between a rope and the ground, you can treat the setup as a right-angle triangle and use the tangent ratio: Tan(θ) = Opposite/Adjacent. Use inverse tangent to find the angle.
How can the cosine ratio be used to solve for an unknown angle?
-To solve for an unknown angle using the cosine ratio, use Cos(θ) = Adjacent/Hypotenuse, rearrange to isolate θ, and then apply the inverse cosine function to find the angle.
Outlines
📐 Introduction to Trigonometric Ratios in Right-Angle Triangles
This paragraph introduces the trigonometric ratios: sine, cosine, and tangent. It explains their use in solving right-angle triangles to find unknown angles or sides. The hypotenuse, opposite, and adjacent sides are defined, and it's emphasized that these ratios only apply to right-angle triangles. The process of labeling sides relative to different angles is described.
🧮 Using Sine to Find the Opposite Side
The sine function is applied to find the length of the opposite side in a right-angle triangle where the hypotenuse and one angle are known. The process involves using the sine of the angle, multiplying by the hypotenuse, and calculating the result using a calculator set in degrees mode. A sample calculation is provided to find the value of the opposite side.
📏 Applying the Cosine Ratio to Find the Hypotenuse
This section demonstrates how to use the cosine ratio to find the hypotenuse of a triangle when given the adjacent side and angle. The cosine formula is explained step by step, including rearranging the equation to isolate the unknown side. A calculation using cosine is performed, and the correct mode for the calculator is emphasized.
✖️ Solving for Sides with Tangent
The tangent function is introduced to find the opposite side when the adjacent side and angle are known. The process involves using the tangent formula and solving for the unknown side. Additionally, it explains how to find angles using the inverse tangent when two sides are known, providing detailed calculations for each scenario.
⛓️ Connecting Trigonometric Ratios with Example Problems
This paragraph presents an example involving the cosine ratio to solve for an adjacent side when the hypotenuse and angle are known. The steps include identifying the hypotenuse and adjacent side, applying the cosine formula, and solving for the unknown adjacent side. The importance of labeling sides correctly is reiterated.
📊 Inverse Trigonometric Functions and Angle Calculation
This section focuses on using inverse trigonometric functions (inverse sine, cosine, and tangent) to calculate angles when two sides of a right-angle triangle are known. The explanation includes how to use the inverse function on a calculator to find the angle, with detailed steps provided for different trigonometric ratios.
🚩 Applying Trigonometric Ratios to Real-World Problems
A real-world example is given where trigonometric ratios are used to find the angle between a flagpole and the ground. The paragraph describes using the tangent ratio with given side lengths to find the angle, followed by a second step to find the length of the rope using the sine or cosine function. The process of solving for the hypotenuse is detailed.
Mindmap
Keywords
💡Right Angle Triangle
💡Hypotenuse
💡Opposite Side
💡Adjacent Side
💡Sine (sin)
💡Cosine (cos)
💡Tangent (tan)
💡SOHCAHTOA
💡Inverse Trigonometric Functions
💡Angle of Elevation
Highlights
Introduction to trigonometric ratios using sine, cosine, and tangent for right-angled triangles.
Explanation of labeling sides of a right-angled triangle: hypotenuse, opposite, and adjacent, based on the angle of interest.
Sine ratio (SOH): sin(θ) equals the opposite side divided by the hypotenuse.
Cosine ratio (CAH): cos(θ) equals the adjacent side divided by the hypotenuse.
Tangent ratio (TOA): tan(θ) equals the opposite side divided by the adjacent side.
Example calculation using sine: solving for an unknown side given an angle of 43° and hypotenuse of 12.
Key note: importance of ensuring calculators are set to degrees mode, not radians, when performing trigonometric calculations.
Example calculation using cosine: solving for the adjacent side when given an angle of 27° and hypotenuse of 5.
Example using tangent: solving for an unknown side using the tangent ratio when given an angle of 38°.
Explanation of inverse trigonometric functions: using the inverse sine (sin⁻¹) to find an angle when the ratio of the opposite to the hypotenuse is known.
Example calculation of an unknown angle using inverse sine (sin⁻¹), yielding an angle of 37°.
Additional example using tangent: solving for an unknown angle using the inverse tangent (tan⁻¹) of 155 divided by 70.
Cosine example applied to real-world problem: calculating the length of a side in a triangle given the hypotenuse and an angle of 42°.
Inverse cosine (cos⁻¹) used to find an unknown angle in a triangle with known adjacent and hypotenuse sides.
Final real-world problem: calculating the angle and length of guy ropes anchored to the ground using tangent and sine ratios.
Transcripts
the trigonometric ratios is talking
about the S ratio the cosine ratio and
the tangent ratio and what it's talking
about is when we have a right angle
triangle we can use S
cosine and tangent to figure out
sometimes an angle within the triangle
and sometimes a side of the triangle if
we're given enough other pieces of
information the trick is to label your
sides correctly first off we must always
have a right angle to be able to apply
this if you have a triangle that's not a
right angle like that or
like that they kind of look the same if
you have a non-right angle triangle you
can't apply this so that's the first
thing you have to check for always
opposite your right angle is the longest
side along the triangle and the longest
side which is opposite that right angle
we call the
hypotenuse the other two are going to be
called the opposite and the adjacent and
that'll depend in relationship to which
angle we're talking about so say our
mystery angle Theta for example is down
here the angle this angle in the
triangle the side that's opposite it is
this one if we spray outwards from that
angle which line do we form we form this
line here so that being opposite our
angle is the opposite side the side
that's left over not already labeled is
the one that's next door to our angle
and because it's next door it's next to
it it's adjacent to it so we call that
the
adjacent side now if if we were talking
about a different angle say this one up
here instead of this one then our sides
would be labeled differently the
hypotenuse would stay the same because
the hypotenuse is always the hypotenuse
it's always that one that's opposite the
right angle but now if we're talking
about this mystery angle up here the
side that's opposite it that it creates
by that angle that angle spays out and
becomes this side down here so now this
this is the opposite and the one left
over the one next door over here would
be the adjacent so that's how we label
our triangle now how do we apply these
ratios the way these ratios work this so
part of the numic that we're using to
remember this soaka TOA stands for the S
of theta equals the
opposite over the hypotenuse so in our
s we have have the S equals the opposite
over the
hypotenuse for K we have the coine
equals the adjacent over the hypotenuse
that's our c ah H and for this part the
TOA we have the tan equals the opposite
over the adjacent that's our to a so for
the S of theta we have the opposite over
the hypotenuse
for the coine of theta we have the
adjacent over the
hypotenuse and for the tangent of theta
we have the
opposite over the
adjacent so say you've been given the
information that the angle down here is
43° and this side which is the
hypotenuse is 12 and trying to find the
length of this side going along here
well because it's a right angle triangle
we can use soaka TOA to work this out
now this x is that the opposite or the
adjacent we look at our angle the piece
of information that we have and we
figure out which line is that angle
creating it's splaying out towards
something and it's creating this line
here so it's opposite that angle which
means this is the opposite this side
over here is the hypotenuse which means
that this one left over is the adjacent
so we don't have any information to do
with the adjacent so we're not applying
that we're going to be using the
opposite and the hypotenuse so we've got
o h so which one of these three are we
going to use we're going to use this one
so we have the S of our mystery angle
43 is equal to the
opposite over the
hypotenuse which
is
our X over 12 now to get X by itself I'm
going to times this 12 over to the other
side or you could call it timesing both
sides by 12 and then this cancels out
either way we get sin 43 * 12 = x so you
just whack that into your calculator and
you need to make sure your calculator is
in degrees mode for this to work out if
your calculator is in Radian
you'll get a different answer so if you
don't know how to set your calculator to
put it in the correct mode make sure you
ask your
teacher for this one we get an answer of
X =
8.18 let's say now that we're given this
angle up here is 27° and we're trying to
find this unknown over here so what
we've got is the hypotenuse over here
the side that is opposite our angle the
side that is created by that angle is
over here so that's the opposite down
there which means that the piece of
information we do have is the adjacent
so we're going to be using the adjacent
and the hypotenuse which we means we've
got ah so which one of these are we
using we're going to be using that the
cosine one so the cosine ratio is that
COS of theta equals the adjacent over
the hyp hpuse which means the COS of 27
is equal to 5 over y now I need to get y
by itself the COS 27 and the Y are
actually just going to swap places but
I'll show you how that works first of
all I need to move the Y away from the 5
so cos 27 * y = 5 now to get the Y by
itself it's been times by the COS 27 so
I need to divide by that so I have y = 5
/ cos 27 and we get an answer of Y =
5.61 say we're given an angle down here
of 38 and we're trying to find this x
the hypotenuse is over here and that
doesn't factor in because that's not one
of the two pieces of information that we
do have so what we have is the side
that's opposite our angle the side that
our angle is forming by spaying out
towards it that's part of it the
opposite and the the other one we've got
is the adjacent so we're dealing with
the opposite and the adjacent we've got
o and a so which one of these are we
using funnily enough we're going to use
tan so we have tan theta equals the
opposite over the adjacent which in this
case would be tan of 38 equals the
opposite X over The adjacent 10 so to
get the X by itself we times the 10 over
here we have tan 38 * * 10 = x so X =
7.8 we can use this if we're trying to
find an angle as well in this case we
have an angle down here our mystery
angle and we've got two of the sides so
let's label them this one over here is
the hypotenuse because it's opposite
that right angle this angle spls out
towards this line going down here so
this three must be the opposite and I
don't need to worry about the adjacent
because it doesn't factor in I don't
have that piece of information I've got
the two I'm going to use so I've got the
opposite and the hypotenuse which means
that I'll be dealing with s so s of
theta is equal to the opposite over the
hypotenuse which means s of theta is
equal to 3 over 5 now how do I work out
Theta if I've got s of theta here well
what we do is we say theta equals the
inverse sign of 3 over5 on your
calculator this looks like a little sign
to the one so it might be the second
function above your sign button you'll
probably press shift sign to get this
little figure s to the 1 that's taking
the inverse sign basically so you do
this with 3.5 in the brackets in your
calculator and you get Theta
equal
37° so we've got our mystery angle up
here this over here let's do our
labeling this would be the hypotenuse
but that's not one of the two pieces of
information we do have so we don't have
to worry about that this angle spays out
towards this line which means that this
down here is the opposite and this is
the one left over it's the next door
neighbor so it's the adjacent so we're
dealing with the opposite and the
adjacent we've got o and a so we're
going to be looking at tan So Tan theta
equals the opposite over the adjacent
which means tan Theta is equal to
155 ided by 70 so to find Theta we take
the inverse tan of 155 over 70 which
gives us Theta equal
66° so in this triangle what is the
length of PQ and remember when they say
two letters like that they're talking
about the line that joins those two
points so we're talking about this line
along here this is line from P to Q so
this is our X down here so the first
thing we need to check is is it a right
angle triangle because if it's not a
right angle triangle then we can't use
soaka TOA but what do you know funnily
enough the example I use in the soak TOA
video is a right angle triangle
so this is our hypotenuse over here
because this is opposite our right angle
what about the other ones here is our
mystery angle and the line that it's
splaying out to create is this one over
here so this would be our opposite which
means our leftover down here is the
adjacent the piece that I'm trying to
find is the adjacent it's this x and the
piece of information that I've been
given is the hypotenuse so I'm dealing
with A and H so in cut TOA which one
gives me an a and an H there it is a h
so I'm using the cosine ratio so I'm
using the COS of theta mystery angle
equals the adjacent over the hypotenuse
and that's where I get my c a h from in
the
C so C of the angle that I have is 42
equals the adjacent side which is my X
over the hypotenuse which is 4
14.5 so to get the X by itself on one
side of the equals so I can solve for x
I'm going to get rid of this 14.5 it's
been divided so I'll times it over here
I have COS of 42 *
14.5 will give me X whack it into your
calculator I get X =
10.78 and it's important to put the
units on there that's going to be
CM what about this one I've got a right
angle triangle so I know I'm going to be
able to use these trigonometric ratios
now which one's the
hypotenuse it's the one that's opposite
the right angle so that's going to be
over here so there's my hypotenuse for
the 6.2 the angle that we're dealing
with is over here and the line that it
creates by splaying out to make a line
the one opposite is over here so this is
our opposite and this must be the next
door neighbor this is the adjacent so
I've got the A and the H again as the
pieces of information so I'm going to be
using cos again COS of theta equal a
over H but this time I'm not solving for
a side I'm solving for an angle itself
so I say COS of theta equals the
adjacent 3.8 over the hypotenuse
6.2 and to find Theta I take the inverse
cos that's like undoing this Co
operation to get it over that side
what's happened to the Theta well we've
kind of caused it so we're going to UNCA
it by taking the inverse CA
of 3.8 over
6.2 I think I've gone off the side of
the screen there sorry whack that into
your calculator and you get
52.2
de last example a flag pole is secured
by Guy ropes anchored to the ground 8 m
from the base of the flag pole and to a
point 9 M up the flag pole find the
angle the gy ropes make with the ground
and the length of the gy ropes okay okay
so the flag pole secured by gy ropes
anchored to the ground 8 m from the base
of the flag pole so these lengths here
are 8 m from the base of the flag pole
in the center so from here to here is
eight and from here to here is eight the
flag pole itself or where these ropes
join up at least that length there whoa
not a straight line is 9
M and do we have a right angle or not
even though it looks like maybe this
triangle isn't
a right angle we can actually assume
that we do have one CU what we're
measuring is off this line right here in
the center at the center of that flag
pole and that makes a right angle going
this way so here's my
triangle and I've got a right angle and
I know two of the sides so first of all
we need to find the angle the guy rope
makes with the ground so that's this
angle there so we're trying to find the
angle between the rope and the ground
which means that's our angle so
hypotenuse comes out from the right
angle that's that one there the opposite
comes out from our mystery angle so
that's that one there and the one that's
left over is the adjacent so that's that
one down there meaning the two pieces of
information that I have the nine and the
eight are the A and the O so I've got O
A I'm going to use tan So Tan of my
mystery angle is equal to the opposite
which is 9 over the adjacent which is 8
so theta equals the inverse tan of 9/ 8
so theta equals
48.4 de now for Part B I'm going to find
the length of the
ropes so now what I'm trying to find is
that length there which is the
hypotenuse so now I'm trying to find the
hypotenuse now I need to be either
finding an angle or at least know one
angle within the triangle I can't use O
A and H because then no angle factors in
so what I'm going to do is the fact that
I've got this piece of information down
here I'll use that as one of the bits of
info that I know and I can either use
the o or the a that is the nine or the
eight to form the ratio so what I mean
by that is I could say the S of
48.4 is equal to the opposite which
would be 9 over the hypotenuse which is
my mystery angle and work it out that
way or using the a instead of the
O I would say the COS of
48.4 is equal to the a which is 8 over
the hypotenuse put either of those into
your calculator and you'll get an answer
of H = 12
m
5.0 / 5 (0 votes)