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# Welcome to Triangles and Solving them - PowerPoint PPT Presentation

Welcome to Triangles and Solving them. This is mainly for higher grade students but foundation students can use the section on Pythagoras’ theorem. Finish. Start. Let’s Get Started. Right-Angled Triangles. Non Right-Angled Triangles. Back. Right-Angled Triangles.

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### Welcome to Triangles and Solving them

This is mainly for higher grade students but foundation students can use the section on Pythagoras’ theorem

Finish

Start

Right-Angled Triangles

Non Right-Angled Triangles

Back

Question has 2 sides and wants the other

Question involves angles

Question involves 3D shapes

Back

This theorem connect the three sides of a right-angle triangle; there are many ways in which it is expressed but here we will use the sequence

If finding the longest side we use the add and if one of the shorter the subtract

In this example, we are finding the shorter side and so will be subtracting

122 - 82 = 144 – 64 = 80

and so the length a is √80 = 8.94 cm

12 cm

8 cm

a

Test questions

Back

x

13 cm

13 cm

x

12 cm

10 cm

Solutions

Back

x

13 cm

13 cm

x

12 cm

10 cm

132+ 122 = 169 + 144

= 313

x is √313 = 17.7 cm

132 - 102 = 169 - 100

= 69

x is √69 = 8.31 cm

Back

In a right-angled triangle where an angle is involved, trigonometry is used and the mnemonic to remember it is SOHCAHTOA. Depending on which two sides are involved, we use one of three rations

Sine = Cosine = and

Tangent

To use we find which two sides are involved and then remember that the button(e.g. sin) is used to find a length and the -1 (e.g. sin-1) for an angle

Formula triangles may be used such as

Hypotenuse

Opposite

a

Cover up then gives the formula such as Hyp = Opp/Sin

Opp

Back

Sin

Hyp

questions

Can I use SOHCAHTOA with angles in right-angled triangles

Q1

Q3. A ship leaves a port and sails 20 km east and then 30 km south. What bearing is the ship from the port

13cm

5cm

x

20km

Q2

x

30km

13°

7cm

Use the -1 to find and angle, button to find length

Q1: Opp =5; hyp =13  Sin-1 (5/13) = 22.6°

Q2: angle = 13° , adj = 7 ; hyp = 7 ÷ Cos 13° = 7.18cm

Q3: opp = 30; adj = 20 : angle = Tan-1(3/2)= 56.3°

Bearing is 90 + 56.3° = 146.3

To Question

This needs one of three formulas depending on what you know

1. Three Sides(SSS) or 2 sides and the angle between(SAS) uses the Cosine rule a2 = b2 + c2 – 2bcCosA2. Any other combination uses the Sine Rule a = b = c

Sin A Sin B Sin C

can be written the other way up if angle is being found

3. Area of triangle = ½ abSinC (again requires SAS)

Cosine rule

Area

Sine rule

Three Sides(SSS) or 2 sides and the angle between(SAS) uses the Cosine rule

a2 = b2 + c2 – 2bcCosA

Can be rewritten as Cos A = b2 + c2 –a2

2bc not on formula sheet

Example

Find the other side (call it a)

a2 = 92 + 82 – 2x9x8xcos82°

= 81 + 64 – 144Cos82°

= 145 – 20

= 125

a = 125 = 11.2cm

If we wanted one of the angles, we would now use the sine rule

8cm

9cm

82°

Find an angle

Can I find sides and angles in non right-angled triangles

Q1 – find the largest angle in a triangle of sides 7cm,8cm and 9cm

Q3. A ship leaves a port and sails 20 km on a bearing of 070 and then 30 km on a bearing 120. How far from the port is it and on what bearing is the ship from the port

Q2

A

B

C

Angle A = 50°, Angle B = 70° and

AC = 12 cm. Find the length of AB

Cosine Rule : a2 = b2 + c2 – 2bcCosA (use for SSS and SAS)

Sine Rule: a = b = c also area of Δ = ½ ab SinC (angle

SinA SinB SinC between)

Q1: cosine rule: largest angle opposite largest side

92 = 7² + 8² - 2x7x8xCosA gives 81= 113-112CosA (not CosA)

Move 113 to give -32=-112CosA so CosA = (-32÷-112) so A=73.4°

Q2: Use Sine rule but note C is 60° so AB ÷ Sin60= 12÷ sin 70 so that AB = 12 x Sin 60 ÷ Sin 70 = 11.1 cm

Q3. from diagram

From question angle PQT=70 and RQT=60 making PQR=130.

Using cosine rule PR² = 20² + 30² -20x30xcos130 = 1686 so PR=41.1km

Using Sine Rule Sin P = 30x Sin 130 ÷ 41.1 so P= 34°

so the bearing of the ship is 104°

Q

70°

70°

60°

P

T

R

To Question

Can I solve 3D Trigonometry Questions

• ABCD is a square of side 7 cm and X is the midpoint of ABCD. M is the midpoint of AD and E is directly above X. Find

• Length EX

• Angle EMX

• Angle ECX

Solve all problems by finding 2D triangles and solving them usually using Pythagoras and SOHCAHTOA

• a. The first step in finding EX is to find AC using the right-angled triangle ADC which will give AC as (7² + 7²) = 9.90. From this AX = ½ of AC = 4.95.

• In Δ EAX we now know EA is 13 and AX = 4.95 so we can find EX using Pythagoras again , EX = (13² - 4.95²) = 12.0 cm

• In Δ EMX for angle EMX, we now know that EX(Opp) is 12.0 and MX(Adj) is 3.5 ( ½ of 7) so that angle is tan-1( 12 ÷ 3.5) = 73.7°

• In Δ ECX for angle ECX , EC(Hyp) = 13 cm and CX(Adj) = 4.95 from part a. This gives us that ECX = Cos-1 ( 4.95 ÷ 13) = 67.6°

To Question

The area of triangle can be found using ½ abSinC

(again requires SAS that is the angle must be between the 2 sides)

A = ½ abSinC

= ½ x 8 x 9 x Sin82

= 35.6 cm²

9 cm

8cm

82°

a = b = c

Sin A Sin B Sin C

Usually written the other way up if finding an angle

A

As finding angle C

Sin C = Sin A

c a

So Sin C = Sin A x c = sin 82 x 6 = 0.74

a 8

So C = sin-1 (0.74)= 48°

6 cm

82°

B

C

8cm

Here are some MyMaths exercises to practice with

Pythagoras

3D trig

SOHCAHTOA

Cosine Rule

angles

Cosine Rule

sides

Sine Rule

Area of triangle

Click anywhere to finish