#26 Trigonometry - Edexcel IGCSE Exam Questions

Summary

In this video, Mr. Aspey walks through several trigonometry problems typically found in the IGCSE exam. He explains step-by-step how to identify the sides of a triangle (opposite, adjacent, hypotenuse) and choose the appropriate trigonometric function (cosine, sine, or tangent) to solve for unknown sides or angles. Throughout the video, Mr. Aspey demonstrates the use of a calculator to perform trigonometric calculations and emphasizes the importance of understanding triangle properties to tackle complex questions. The video concludes with a detailed solution to a perimeter problem involving multiple triangles.

Takeaways

• 📐 Label the sides of the triangle correctly by identifying the opposite, adjacent, and hypotenuse based on the angle and right angle.
• 🧮 To find PQ, use the cosine rule: cos(angle) = adjacent/hypotenuse, then calculate hypotenuse using the formula H = A/cos(angle).
• 📊 For the second triangle, to find EG, use cosine again: cos(angle) = adjacent/hypotenuse. Apply the formula to calculate the side length.
• ✖️ To solve the second triangle, use the known side length and angle to calculate the adjacent side using tan: tan(angle) = opposite/adjacent.
• 📏 For more complex triangles, break down the larger shape into smaller triangles to solve unknown sides or angles using trigonometric identities.
• 🔺 Use the inverse of tan (tan-1) when solving for an unknown angle with known opposite and adjacent sides.
• 📉 Work out multiple steps by finding intermediate side lengths and angles in small triangles before solving the larger figure.
• 🧩 Drawing extra lines (like perpendiculars) can help simplify complex trapezium problems by creating right-angle triangles.
• 📏 Use trigonometric functions like sine, cosine, and tangent to solve for missing sides and angles when working with trapeziums.
• 📝 For perimeter questions involving repeated shapes, calculate individual side lengths and sum them up, accounting for repeated sides.

Q & A

What is the first step in solving a trigonometry problem involving right-angled triangles?

The first step is to label the sides of the triangle. You need to identify the 'opposite', 'adjacent', and 'hypotenuse' sides relative to the given angle.

Which trigonometric ratio is used when you know the adjacent and need to find the hypotenuse?

The cosine ratio is used when you know the adjacent side and need to find the hypotenuse. The formula is cos(angle) = adjacent/hypotenuse.

How do you solve for the hypotenuse if you know the adjacent side and the angle?

To solve for the hypotenuse (H), use the formula H = adjacent / cos(angle). Substitute the known values into this equation.

In the script, how is PQ calculated?

PQ is calculated by using the cosine rule: PQ = 24.3 / cos(63). When entered into a calculator, it results in 53.5.

What trigonometric ratio is used when finding a side opposite to a known angle?

The sine ratio is used when finding a side opposite a known angle. The formula is sin(angle) = opposite/hypotenuse.

How is the adjacent side (A) found when the opposite and angle are known?

To find the adjacent side (A) when the opposite is known, use the tangent formula: A = opposite / tan(angle).

How is the angle in a triangle calculated if you know the opposite and adjacent sides?

The angle is calculated using the inverse tangent function (tan^-1). The formula is angle = tan^-1(opposite/adjacent).

In the perimeter problem of the trapezium, what was the process for finding the total perimeter?

The total perimeter was calculated by finding each side of the trapezium using trigonometry (adjacent and hypotenuse) and adding them together. The final perimeter was 101.4 cm.

How is the hypotenuse calculated in the problem involving the shape with five triangles?

The hypotenuse was calculated using Pythagoras' theorem: hypotenuse^2 = opposite^2 + adjacent^2. The final hypotenuse was 13.5 units.

What formula is used to calculate the total perimeter of the shape with five triangles?

To calculate the total perimeter, the lengths of the five hypotenuses and five short sides were added together. The total perimeter was calculated as 111 units.

Outlines

00:00

📐 Introduction to Trigonometry Problem-Solving

In this section, Mr. Aspey introduces the topic of trigonometry, focusing on labeling the sides of a triangle based on the given angle. He identifies the adjacent, opposite, and hypotenuse and explains that cosine is the appropriate trigonometric ratio to use for the first problem. By substituting values into the cosine formula, Mr. Aspey calculates PQ using 24.3 as the adjacent and 63 degrees as the angle, yielding 53.5 as the result. He then transitions to working on another triangle to calculate side lengths.

05:01

📊 Using Cosine to Solve for Triangle Sides

This section continues with another triangle, where Mr. Aspey introduces a second problem. He starts by identifying the sides of the triangle and calculating the adjacent side (Y) using cosine. Substituting the values for the angle (40 degrees) and the hypotenuse (12), he calculates Y as 9.19. With this value, he moves to another triangle and identifies the opposite and adjacent sides. Using the tangent function, Mr. Aspey determines the adjacent side as 17.3 cm by dividing the opposite (9.19) by tan(28 degrees).

10:04

🔺 Finding Angles Using Trigonometry

In this part, Mr. Aspey shifts focus to finding an unknown angle within a triangle. He first identifies the opposite and adjacent sides and uses the tangent function to solve for the angle. He inputs tan-1(12.73/13) into his calculator, yielding 44.4 degrees for the total angle. He then subtracts 20 degrees to isolate the angle needed, arriving at 24.4 degrees. This segment demonstrates how to find angles by working backward using inverse trigonometric functions.

📏 Calculating Perimeter Using Trigonometry and Geometry

Mr. Aspey tackles a challenging problem where he is asked to calculate the perimeter of a trapezium. To solve it, he creates a right triangle by drawing a perpendicular line and identifies the sides of the triangle. Using the tan function, he finds the adjacent side as 17.9. He then uses the sine function to find the hypotenuse (Y), which comes out to be 24.5. Finally, Mr. Aspey calculates the total perimeter of the trapezium by adding up the known side lengths and reaches 101.4 cm.

🔻 Perimeter of Composite Shapes

The final part involves calculating the perimeter of a complex shape made of five smaller triangles. Mr. Aspey finds the hypotenuse and adjacent sides using tan and Pythagoras’ theorem. After calculating the hypotenuse (13.5) and adjacent (4.16), he calculates the total perimeter by summing the sides. The total perimeter is 111 cm after multiplying and subtracting the appropriate values. He concludes the video by encouraging viewers to like and subscribe for more trigonometry problem-solving content.

Keywords

💡Trigonometry

Trigonometry is the branch of mathematics that studies relationships between the angles and sides of triangles. In the video, trigonometry is used to solve various problems involving right-angled triangles, calculating unknown side lengths and angles using trigonometric ratios such as sine, cosine, and tangent.

💡Cosine

Cosine is a trigonometric function used to relate the angle of a triangle to the ratio of the adjacent side to the hypotenuse. In the script, the cosine function is employed to calculate side lengths when the angle and other side lengths of a right-angled triangle are known, as seen when the teacher solves for 'H' in the triangle.

💡Tangent

Tangent (tan) is another trigonometric function that relates an angle of a triangle to the ratio of the opposite side over the adjacent side. The video frequently uses tangent to find unknown sides or angles in right-angled triangles, such as when solving for side lengths or angles in multiple triangle setups.

💡Hypotenuse

The hypotenuse is the longest side of a right-angled triangle, opposite the right angle. In trigonometry, the hypotenuse plays a crucial role in applying sine and cosine ratios. The teacher repeatedly refers to the hypotenuse as part of the calculations when solving for other side lengths in various triangles.

💡Adjacent

Adjacent refers to the side of a right-angled triangle that is next to a given angle and forms part of the angle along with the hypotenuse. The video uses the adjacent side in relation to cosine and tangent functions to solve for unknown side lengths in several problems.

💡Opposite

Opposite refers to the side of the triangle that is directly across from a given angle. In trigonometry, this side is important when applying the sine and tangent ratios. The teacher points out the opposite side frequently when labeling the triangle to prepare for solving for unknowns.

💡Perimeter

The perimeter is the total distance around the edges of a shape, such as a triangle or trapezium. In the script, the teacher calculates the perimeter of a trapezium and a composite shape, adding up all the side lengths, including those found using trigonometric principles.

💡Pythagoras' Theorem

Pythagoras' Theorem is a fundamental principle in geometry that relates the sides of a right-angled triangle: a² + b² = c², where c is the hypotenuse. The teacher references this theorem as an alternative to using trigonometry to find the hypotenuse when two sides of a triangle are known.

💡Angle

An angle is the space between two intersecting lines or surfaces, measured in degrees. In trigonometry, angles are essential to calculating unknown sides using trigonometric functions like sine, cosine, and tangent. Throughout the video, angles like 63°, 40°, and 20° are key to solving various triangle problems.

💡Calculator

A calculator is an electronic tool used to perform mathematical operations quickly, such as arithmetic, trigonometric functions, and fractions. The teacher frequently instructs viewers to use their calculator to compute trigonometric values like 'cosine of 63' or 'tan of 28,' demonstrating the importance of this tool in solving IGCSE trigonometry questions.

Highlights

Introduction to labeling the sides of a right triangle based on the given angle, including opposite, hypotenuse, and adjacent.

Use of cosine rule to find the hypotenuse by covering the relevant part of the triangle formula and calculating it with a calculator.

Demonstration of finding the side length EG using cosine, and breaking down the triangle into smaller components.

Explanation of solving for side length Y using cosine and then applying that to solve for another side in the second triangle.

Use of the tan rule to calculate the adjacent side in a triangle based on the opposite side and angle.

Step-by-step method for solving a more complex triangle problem by combining known sides and angles to find unknowns.

Finding angle Theta using the inverse of tan with the known opposite and adjacent sides.

Subtracting 20 degrees from the calculated total angle to get the desired internal angle for the problem.

Breaking down a trapezium into right triangles and rectangles to calculate unknown side lengths.

Applying tan to find adjacent sides in the trapezium problem by using the opposite side and given angle.

Use of sine to solve for the hypotenuse in a right triangle based on the known opposite side and angle.

Summing the perimeter of a complex shape by calculating the individual side lengths and adding them together.

Breaking down a five-sided figure into smaller triangles to find the perimeter, using a combination of tan and Pythagoras’ theorem.

Final calculation of the perimeter of the five-sided shape by multiplying the sides and adding the results.

Concluding the session by summarizing the key points and encouraging viewers to like and subscribe.

Transcripts

hello my name is Mr aspey and this is

trigonometry as part of my IGCSE exam

question series if you do find this

video useful please do like and

subscribe now let's get into the

maths okay first thing to do is to label

the sides so there's the angle so this

is the opposite there's the right angle

so that's the hypotenuse and this

therefore must be the

adjacent now the sides that we're

looking for are uh PQ which is this one

here and we have the adjacent so A and H

means I need to use cosine so I draw my

triangle and I write cosine of the angle

is a over H and because I'm looking for

H it means I cover up that part of the

triangle and that tells me that H is

equal to a over cosine of the

angle so substituting in H

is uh PQ which is what we're looking for

and a is

24.3 and that's over cosine of the angle

which is

63 so I go to my calculator and I just

type in fraction button

24.3 over cine of 63 close brackets

equals and it gives me 53.5

uh next question I need to work out e g

which is this side here let's call it

x uh and if I look at that triangle um I

don't have much to go on really I've

just got the side I'm looking for an

angle but I don't have another side so I

need to look at this triangle here in

order to work out the side length e

which will call

Y and once I know that then I can use

the second

triangle okay so let's label the sides

we have opposite the

angle we have opposite the right angle

is the

hypotenuse and we have here is the

adjacent the sides that I

have is H and the side I'm looking for

is a so that's

cosine so I draw triangle and I write

cosine of the angle is a over H I'm

looking for a so I cover a and that

means that I need to multiply the two at

the bottom to get a so I write a is

equal to cosine of the angle Time by H

so in this case our a is we labeled y

cosine of the angle is cosine of 40

multiplied by H which is 12 so I go to

my calculator and I type cine of 40

close brackets Times by 12 and that's

equal to

9.12 sorry

9.19 okay now I can look at my um my

second

triangle over here I know what this is

now

9.19 and I have the angle so this is now

the

opposite and there's the right angle so

this is the

hypotenuse and finally this is the

adjacent so the side that I have is the

opposite and the side that I'm looking

for is the

adjacent so that means I need to use the

tan

triangle o and

a and I'm looking for a so I cover the

adjacent and that tells me if I want to

work out a I need to do the opposite

over

tan of the

angle the opposite is

9.19 and we have tan of

28 so that goes into our

calculator I can press fraction button I

can press answer because that is stored

the last answer which is 9.19 and I can

go tan of 28 close brackets equals

17. 28 so 17

.3 and that is

CM okay um slightly tricky question here

I need to work out this angle in

here um and in order to do that I'm

going to need to know what the base and

the height of this overall triangle is

and at the moment that is eight is what

I know so it' be very useful if I was to

know this let's call it X in here so I

can use the smaller triangle to order to

work out that

X um there's the angle so that's the

opposite opposite the right angle is

hypotenuse and this is the

adjacent so the side length which I'm

looking for is the opposite and the one

that I have is the adjacent so I will

use tan

and I write tan of uh the angle is equal

to o

a and I'm looking for o so I cover that

so to find the opposite here I do tan of

the angle multip

a and the opposite we've labeled X and

we have tan of the angle 20 * by

13 so we go into our calculator and we

do

tan of 20 close brackets Times by

133 and that gives me

4.73 okay that's good now what I'm able

to do is look at the overall

triangle and that

triangle looks like this 13 and 8 + 14

uh 8 + 4.7 is

12.7 uh down that side there

so I could work out this angle in here

let's call it

Theta and I have the opposite and the

adjacent so I'm going to be using tan

again but this time I am looking for the

angle so I cover tan

Theta and that tells me that tan Theta

is equal to the opposite which is 12.

73 over the adjacent which is 13

so on my calculator to work out Theta I

would do

tan-1 of

12. 73 over

13 and that gives me

44.4 but we're not quite finished

because B

A is just this angle in

here and I've worked out the total angle

so I need to take away 20 in order to

get the bit that I actually

need so that would be

24.4

perfect okay next question we're asked

to work out the perimeter of the

trapezium so I know this side and I know

this side uh but I don't know this side

or this side so that's a bit of an

issue um so what I need to do and what

makes this question quite tricky is I

got to draw My Own Line in so I'm going

to draw a perpendicular down there to

create a right angle triangle and a

rectangle now of course this rectangle

on the bottom would be uh 12 21.2 up to

here so now what I'll need to do is I'll

need to work out um let's call This One

X and this one

y uh I have an angle and I need another

side and that side is going to be 16.7

which is going to be this height here

that side there is the

opposite this side is the hypotenuse and

this is the

adjacent so let's first try and find the

adjacent and I will use tan to do

that so I'll do tan of the angle o over

a and I'm looking for the adjacent so I

cover the

adjacent and that tells me that X the

adjacent is equal to the opposite

16.7 over the

hypotenuse sorry opposite over tan of

the

angle so that goes into my calculator

16.7

over tan of the

angle and I get

17.9 okay now how can I work out why

well I could use Pythagoras cuz now I've

got two sides but seeing this is a

trigonometry video I'm going to use

trigonometry uh and I'm going to use

sign so I don't think I've use sign yet

so s of theta and then we've got o over

H I'm looking for the

hypotenuse so I cover up the

hypotenuse so that tells me the

hypotenuse which I've labeled Y is equal

to the opposite

16.7

over tan sorry s

of the angle

43 so I would do uh

16.7

over uh sine of

43 and that gives me

24.5 so the total

perimeter is equal to 21.2 * 2 cuz there

are two of them plus a

16.7 there plus X and Y which is

17.9 and

24.5 so I do 24.5 I'll Plus on

17.9 I will Plus on 16.7 and I will Plus

on

21.2

twice and I'll get

101.5 so three significant figures

actually that's 101.4

the four won't round up so it's just

101

cm and here we have a tricky question so

you've got five of these shapes and it's

asking you to work at the perimeter of

the total shape um so what I'm going to

do first is I'm going to find all the

sides of this triangle so the side that

I do know is the

opposite so let's try and find out the

uh

hypotenuse and the

adjacent uh I'll start with the adjacent

which means I have the O and I'm looking

for the a which means I need to use

tan So Tan is O over

a and I'm looking for the a so I'm going

to have to do um the

opposite which is 12.8 over tan of 72

uh 12.8

over tan of

72 and that gives me

4.15 oh sorry

4.16 and next we can find the

hypotenuse and I will do that by uh

using p

theorem so that would be < TK of 12.8

SAR plus the last answer

squared and that will be

13.5 so that's

13.5 and this one is

4.16 okay now we can look

at um the hypotenuse sorry the um the

perimeter so we have have five of

these and those are all the um uh 12.8

so 5 *

12.8 and we also

have uh 1 2

3 4 five of these and they are all the

hypotenuse so 5 * the hypotenuse

13.5

minus

4.16 okay so let's just put into our

calculator and that's going to give us

the answer so it's 5 *

12.8 plus 5

* 13.5 which is the hypotenuse minus the

short side and that's going to give us

that purple

distance and we get an answer of 11.7

which to three significant figures is

111 and we're

done thanks for watching if you found

that useful please do like and subscribe

and watch the next video in the series

bye for now