Maths Tutorial: Trigonometry SOH CAH TOA (trigonometric ratios)

Summary

The 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

00:00

📐 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.

05:02

🧮 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.

10:02

📏 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.

15:03

✖️ 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.

Keywords

💡Right Angle Triangle

A right angle triangle is a triangle where one of the angles is exactly 90°. This is crucial in trigonometry because trigonometric ratios like sine, cosine, and tangent only apply to right-angle triangles. In the video, the narrator emphasizes that without a right angle, these ratios can't be used, setting a foundation for calculating angles and sides in such triangles.

💡Hypotenuse

The hypotenuse is the longest side of a right-angle triangle, located opposite the right angle. It plays a critical role in trigonometric ratios such as sine and cosine, which compare the hypotenuse with other sides. In the video, the hypotenuse is repeatedly referenced when calculating unknown sides or angles, demonstrating its importance in trigonometric calculations.

💡Opposite Side

The opposite side refers to the side of a right-angle triangle that is opposite to the angle being considered. It is essential in the sine and tangent trigonometric ratios, which involve the opposite side in relation to either the hypotenuse or the adjacent side. In the video, it is shown how the opposite side shifts depending on the angle under consideration.

💡Adjacent Side

The adjacent side is the side next to the angle being considered, excluding the hypotenuse. It plays a role in the cosine and tangent trigonometric ratios. The video explains that identifying the adjacent side is important when working with angles, as it helps determine which trigonometric ratio to use.

💡Sine (sin)

Sine is a trigonometric function that relates an angle in a right-angle triangle to the ratio of the opposite side over the hypotenuse. It is represented as sin(θ) = Opposite/Hypotenuse. In the video, sine is used to calculate unknown side lengths when the angle and one other side are known.

💡Cosine (cos)

Cosine is another trigonometric function that relates an angle to the ratio of the adjacent side over the hypotenuse, expressed as cos(θ) = Adjacent/Hypotenuse. The video illustrates how cosine can be used to find unknown sides or angles, particularly when the hypotenuse and adjacent sides are given.

💡Tangent (tan)

Tangent is a trigonometric function that represents the ratio of the opposite side to the adjacent side in a right-angle triangle, shown as tan(θ) = Opposite/Adjacent. The video demonstrates its use in problems where only the opposite and adjacent sides are known, allowing for the calculation of angles.

💡SOHCAHTOA

SOHCAHTOA is a mnemonic used to remember the trigonometric ratios: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, and Tangent = Opposite/Adjacent. In the video, the narrator repeatedly refers to SOHCAHTOA to guide the viewer through solving trigonometric problems with right-angle triangles.

💡Inverse Trigonometric Functions

Inverse trigonometric functions, such as inverse sine (sin⁻¹), inverse cosine (cos⁻¹), and inverse tangent (tan⁻¹), are used to calculate an angle when the sides of a triangle are known. In the video, the narrator explains how to find angles by using these inverse functions, particularly in cases where side lengths are given and an angle is unknown.

💡Angle of Elevation

An angle of elevation is the angle formed by the horizontal ground and the line of sight to an object above the ground. In the video, the concept is applied when discussing how to calculate the angle made by a guy rope securing a flagpole to the ground, using trigonometric functions to determine the angle between the rope and the ground.

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