Trigonometry (4103).

Trigonometry (4103)

Trigonometry “triangle measure”

A little bit of review...

The 3 angles from a triangle ALWAYS equal 180o
b c a + b + c = 180o

Find the total of the other angles
30◦

= 90◦ 30◦

Total angles = 180◦ 90◦ + 30◦ + a = 180◦ 120◦ + a = 180◦ a = 180◦ – 120◦ a = 60◦ a = 90◦ 30◦

All sides are the same length
Equilateral triangle All sides are the same length

Equilateral triangle All angles are the same (180o ÷ 3 = 60o)

Two sides are the same length
Isosceles triangle Two sides are the same length

Isosceles triangle Two angles are the same

No sides are the same length
Scalene triangle No sides are the same length

Scalene triangle No angles are the same

Right-angled triangle
hypotenuse side side

side opposite to angle A hypotenuse A side adjacent (next to) angle A

hypotenuse (c) side (a) 90o side (b)

Pythagorean Theorem c2 = a2 + b2 hypotenuse (c) side (a) side (b)

What if you switch a and b?
c2 = a2 + b2 hypotenuse (c) side (a) side (b)

What if you switch a and b?
c2 = a2 + b2 hypotenuse (c) side (b) Doesn’t matter, they’re both sides! side (a)

B side adjacent to angle B hypotenuse A side opposite to angle B

What is the length of the hypotenuse?
c2 = a2 + b2 hypotenuse (c) side (a) x cm 3 cm side (b) 4 cm

What is the length of the hypotenuse?
c2 = a2 + b2 hypotenuse (c) x2 = x2 = x2 = 25 x2 = 25 x = 5 cm side (a) x cm 3 cm side (b) 4 cm

What is the length of the side?
c2 = a2 + b2 hypotenuse (c) side (a) 10 cm x cm side (b) 5 cm

What is the length of the side?
c2 = a2 + b2 102 = x2 + 52 100 = x2 + 25 100 – 25 = x2 75 = x2 x2 = 75 x = 8.7 cm hypotenuse (c) side (a) 10 cm x cm side (b) 5 cm

Trigonometric ratios depend on which angle is used
sine cosine tangent

Sine ratio (SOH) sin A = opposite hypotenuse A side opposite
to angle A hypotenuse A side adjacent to angle A

Sine ratio (SOH) sin B = opposite hypotenuse B A side adjacent
to angle B hypotenuse A side opposite to angle B

Cosine ratio (CAH) cos A = adjacent hypotenuse A side opposite

Cosine ratio (CAH) cos B = adjacent hypotenuse B A side adjacent

Tangent ratio (TOA) tan A = opposite adjacent A side opposite

Tangent ratio (TOA) tan B = opposite adjacent B A side adjacent

SOH CAH TOA Trigonometric ratios sin θ = opp cos θ = adj tan θ = opp
hyp hyp adj

Find the lengths of the missing sides and angle (right triangle)
B 7 cm 35o A C

Find the lengths of the missing sides and angle (right triangle)
B 7 cm 35o 90o A C

Step 1. List the information given, and what is needed
What we know: mBC = 7 cm A = 35o C = 90o B 7 cm What we need: mAB = ? mAC = ? B = ? 35o 90o A C

Step 2. Find the missing side AB
Look at the triangle from A: mBC = opposite mAB = hypotenuse B hyp 7 cm (opp) 35o 90o A C

Look at the triangle from A: mBC = opposite mAB = hypotenuse B hyp 7 cm (opp) ? = opp 35o 90o hyp A C

Look at the triangle from A: mBC = opposite mAB = hypotenuse B hyp 7 cm (opp) sin θ = opp 35o 90o hyp A C

sin θ = opp B hyp hyp sin 35o = opp 7 cm (opp) hyp 35o 90o A C

sin θ = opp B hyp hyp sin 35o = opp 7 cm (opp) hyp 0.574 = 7 cm 35o 90o hyp A C

sin θ = opp B hyp hyp sin 35o = opp 7 cm (opp) hyp 0.574 = 7 cm 35o 90o hyp A C 0.574 (hyp) = 7 cm

sin θ = opp B hyp hyp sin 35o = opp 7 cm (opp) hyp 0.574 = 7 cm 35o 90o hyp A C 0.574 (hyp) = 7 cm hyp = 12.2 cm

Step 3. Find the missing side AC
Look at the triangle from A: mBC = opposite mAB = hypotenuse mAC = adjacent B 12.2 cm (hyp) 7 cm (opp) 35o 90o A C (adj)

Look at the triangle from A: mBC = opposite mAB = hypotenuse mAC = adjacent B 12.2 cm (hyp) 7 cm (opp) Since we have two sides, we have a choice of trig ratios! 35o 90o A C (adj)

cos θ = adj B hyp 12.2 cm (hyp) or 7 cm (opp) tan θ = opp 35o 90o adj A C (adj)

cos θ = adj B hyp 12.2 cm (hyp) cos 35o = adj 7 cm (opp) hyp 35o 90o A C (adj)

cos θ = adj B hyp 12.2 cm (hyp) cos 35o = adj 7 cm (opp) hyp 0.819 = adj 35o 90o 12.2 cm A C (adj)

cos θ = adj B hyp 12.2 cm (hyp) cos 35o = adj 7 cm (opp) hyp 0.819 = adj 35o 90o 12.2 cm A C (adj) 0.819 (12.2 cm) = adj

cos θ = adj B hyp 12.2 cm (hyp) cos 35o = adj 7 cm (opp) hyp 0.819 = adj 35o 90o 12.2 cm A C (adj) 0.819 (12.2 cm) = adj adj = 10 cm

tan θ = opp B adj 12.2 cm (hyp) tan 35o = opp 7 cm (opp) adj 35o 90o A C (adj)

tan θ = opp B adj 12.2 cm (hyp) tan 35o = opp 7 cm (opp) adj 0.700 = 7 cm 35o 90o adj A C (adj)

tan θ = opp B adj 12.2 cm (hyp) tan 35o = opp 7 cm (opp) adj 0.700 = 7 cm 35o 90o adj A C (adj) 0.700 (adj) = 7 cm

tan θ = opp B adj 12.2 cm (hyp) tan 35o = opp 7 cm (opp) adj 0.700 = 7 cm 35o 90o adj A C 10 cm (adj) 0.700 (adj) = 7 cm adj = 10 cm

Step 4. Find the missing angle B
12.2 cm (hyp) 7 cm (opp) 35o 90o A C 10 cm (adj)

180o = A + B + C B 12.2 cm (hyp) 7 cm (opp) 35o 90o A C 10 cm (adj)

180o = A + B + C B 180o = 35o + B + 90o 12.2 cm (hyp) 7 cm (opp) 35o 90o A C 10 cm (adj)

180o = A + B + C B 180o = 35o + B + 90o 12.2 cm (hyp) 180o = B + 125o 7 cm (opp) 35o 90o A C 10 cm (adj)

180o = A + B + C B 180o = 35o + B + 90o 12.2 cm (hyp) 180o = B + 125o 7 cm (opp) B = 180o – 125o 35o 90o A C 10 cm (adj)

180o = A + B + C B 180o = 35o + B + 90o 12.2 cm (hyp) 55o 180o = B + 125o 7 cm (opp) B = 180o – 125o 35o 90o B = 55o A C 10 cm (adj)

Steps to completing a right triangle
Step 1. List the information given, and what is needed Step 2. Find the missing side(s) Step 3. Find the missing angle(s)

Find the length of the missing side and angles
25 cm 19 cm 30o B C

Step 1. List the missing information, and what is needed
What we know: mAB = 25 cm mAC = 19 cm B = 30o A 25 cm 19 cm What we need: mBC = ? A = ? C = ? 30o B C

Step 2. Create a 90o angle by cutting the triangle in two
Start at the top angle and continue until it hits the bottom of the triangle at a 90o angle 25 cm 19 cm 30o B H C

Name the point of intersection H A 25 cm 19 cm 30o B H C

Name the point of intersection H A 25 cm 19 cm Now find the missing information for each new triangle! 30o B H C

Step 3. Find the length BH Look at the new triangle from B:
mBH = adjacent mAB = hypotenuse A 25 cm 19 cm 30o B H C

Step 3. Find the length BH ? = adj hyp
Look at the new triangle from B: mBH = adjacent mAB = hypotenuse A 25 cm 19 cm ? = adj hyp 30o B H C

Step 3. Find the length BH cos θ = adj hyp
Look at the new triangle from B: mBH = adjacent mAB = hypotenuse A 25 cm 19 cm cos θ = adj hyp 30o B H C

Step 3. Find the length BH cos θ = adj hyp cos 30o = adj hyp A 25 cm

Step 3. Find the length BH cos θ = adj hyp cos 30o = adj hyp
25 cm 19 cm hyp 0.866 = adj 25 cm 30o B H C

25 cm 19 cm hyp 0.866 = adj 25 cm 30o (0.866)(25 cm) = adj B H C

25 cm 19 cm hyp 0.866 = adj 25 cm 30o (0.866)(25 cm) = adj B 21.7 cm H C adj = 21.7 cm

Step 4. Find the angle A 180o = A + B + H A 25 cm 19 cm 30o B H C

Step 4. Find the angle A 180o = A + B + H 180o = A + 30o + 90o A 25 cm

Step 4. Find the angle A 180o = A + B + H 180o = A + 30o + 90o
25 cm 19 cm 30o B 21.7 cm H C

25 cm 19 cm A = 180o – 120o 30o B 21.7 cm H C

25 cm 19 cm A = 180o – 120o A = 60o 30o B 21.7 cm H C

Step 5. Find the length AH There are many different ways to find mAH:
– Pythagoras – tan A or tan B – cos A – sin B A 60o 25 cm 19 cm 30o B 21.7 cm H C

Step 5. Find the length AH sin θ = opp hyp A 60o 25 cm 19 cm 30o B H C

Step 5. Find the length AH sin θ = opp hyp sin 30o = opp hyp A 60o
25 cm 19 cm hyp 30o B 21.7 cm H C

Step 5. Find the length AH sin θ = opp hyp sin 30o = opp hyp
25 cm 19 cm hyp 0.500 = opp 25 cm 30o B 21.7 cm H C

25 cm 19 cm hyp 0.500 = opp 25 cm 30o (0.500)(25) = opp B 21.7 cm H C

25 cm 19 cm hyp 12.5 cm 0.500 = opp 25 cm 30o (0.500)(25) = opp B 21.7 cm H C hyp = 12.5 cm

Step 6. Find the angle C ? = opp hyp Look at the new triangle from C:
mAH = opposite mAC = hypotenuse A 60o 25 cm 19 cm 12.5 cm ? = opp hyp 30o B 21.7 cm H C

Step 6. Find the angle C sin θ = opp hyp
Look at the new triangle from C: mAH = opposite mAC = hypotenuse A 60o 25 cm 19 cm 12.5 cm sin θ = opp hyp 30o B 21.7 cm H C

Step 6. Find the angle C sin θ = opp hyp A 60o 25 cm 19 cm 30o B H C
12.5 cm 30o B 21.7 cm H C

Step 6. Find the angle C sin θ = opp hyp sin θ = opp hyp A 60o 25 cm
12.5 cm 30o B 21.7 cm H C

Step 6. Find the angle C sin θ = opp hyp sin θ = opp hyp
25 cm 19 cm sin θ = 12.5 cm 12.5 cm 19 cm 30o B 21.7 cm H C

25 cm 19 cm sin θ = 12.5 cm 12.5 cm 19 cm sin θ = 0.66 30o B 21.7 cm H C

25 cm 19 cm sin θ = 12.5 cm 12.5 cm 19 cm sin θ = 0.66 30o sin-1(0.66) = θ B 21.7 cm H C

25 cm 19 cm sin θ = 12.5 cm 12.5 cm 19 cm sin θ = 0.66 30o 41.1o sin-1(0.66) = θ B 21.7 cm H C θ = 41.1o

Step 7. Find the length CH There are many different ways to find mCH:
– Pythagoras – cos C – tan C A 60o 25 cm 19 cm 12.5 cm 30o 41.1o B 21.7 cm H C

Step 7. Find the length CH cos θ = adj hyp A 60o 25 cm 19 cm 30o B H C
12.5 cm 30o 41.1o B 21.7 cm H C

Step 7. Find the length CH cos θ = adj hyp cos 41.1o = adj hyp A 60o
25 cm 19 cm 12.5 cm 30o 41.1o B 21.7 cm H C

Step 7. Find the length CH cos θ = adj hyp cos 41.1o = adj hyp
25 cm 19 cm 12.5 cm 0.754 = adj 19 cm 30o 41.1o B 21.7 cm H C

25 cm 19 cm 12.5 cm 0.754 = adj 19 cm 30o 41.1o 0.754 (19 cm) = adj B 21.7 cm H C

25 cm 19 cm 12.5 cm 0.754 = adj 19 cm 30o 41.1o 0.754 (19 cm) = adj B 21.7 cm H 14.3 cm C adj = 14.3 cm

Step 8. Find the angle A 180o = A + C + H A 60o 25 cm 19 cm 30o B H C
12.5 cm 30o 41.1o B 21.7 cm H 14.3 cm C

Step 8. Find the angle A 180o = A + C + H 180o = A + 41.1o + 90o A 60o
25 cm 19 cm 12.5 cm 30o 41.1o B 21.7 cm H 14.3 cm C

Step 8. Find the angle A 180o = A + C + H 180o = A + 41.1o + 90o
25 cm 19 cm 12.5 cm 30o 41.1o B 21.7 cm H 14.3 cm C

25 cm 19 cm A = 180o – 131.1o 12.5 cm 30o 41.1o B 21.7 cm H 14.3 cm C

25 cm 19 cm A = 180o – 131.1o 12.5 cm A = 48.9o 30o 41.1o B 21.7 cm H 14.3 cm C

Step 9. Complete triangle
180o = A + B + C A = 60o o A = 108.9o 48.9o 60o 25 cm 19 cm 12.5 cm 30o 41.1o B 21.7 cm H 14.3 cm C

180o = A + B + C A = 60o o A = 108.9o 108.9o 180o = 108.9o + 30o o 25 cm 19 cm 30o 41.1o B 21.7 cm H 14.3 cm C

180o = A + B + C A = 60o o A = 108.9o 108.9o 180o = 108.9o + 30o o 25 cm 19 cm The angles in the original triangle ABC add up to 180o 30o 41.1o B 21.7 cm H 14.3 cm C

mBC = mBH + mCH A 108.9o 25 cm 19 cm 30o 41.1o B 21.7 cm H 14.3 cm C

mBC = mBH + mCH A 108.9o mBC = 25 cm 19 cm 30o 41.1o B 21.7 cm H 14.3 cm C

mBC = mBH + mCH A 108.9o mBC = mBC = 36 cm 25 cm 19 cm 30o 41.1o B 36 cm C

Steps to complete a non-right angle triangle
Step 1. List the missing information, and what is needed Step 2. Create 90o angles by cutting the triangle in two Step 3. Looking at the first triangle, solve for missing angle(s) and/or side(s) Step 4. Looking at the second triangle, solve for missing angle(s) and/or side(s) Step 5. Put the halves of sides and angles together into the one original triangle

So far, there are two ways to solve a right-angled triangle:

So far, there are two ways to solve a right-angled triangle:
Pythagoras (c2 = a2 + b2) Trigonometric ratios (SOH CAH TOA)

Isn’t there another way to solve a non-right angled triangle?

Isn’t there another way to solve a non-right angled triangle?
Yes! Sin Law and Cos Law

Sine Law Uses the sine ratio

Sine Law a = b = c sin A sin B sin C

Sine Law lengths a = b = c sin A sin B sin C angles

25 cm 19 cm 30o B C

Remember: Capital letters = angles Lower-case letters = sides A c b 25 cm 19 cm 30o B a C

angle A ↔ side a Remember: Capital letters = angles Lower-case letters = sides A c b 25 cm 19 cm Angles and sides with the same letters are opposite each other 30o B a C

angle B ↔ side b Remember: Capital letters = angles Lower-case letters = sides A c b 25 cm 19 cm Angles and sides with the same letters are opposite each other 30o B a C

angle C ↔ side c Remember: Capital letters = angles Lower-case letters = sides A c b 25 cm 19 cm Angles and sides with the same letters are opposite each other 30o B a C

What we know: mAB = c = 25 cm mAC = b = 19 cm B = 30o A c b 25 cm 19 cm What we need: mBC = a = ? A = ? C = ? 30o B a C

Step 2. Find one ‘pair’, and use it to fill in another ‘pair’
We have both angle B and side b A c b 25 cm 19 cm We can use these to fill out the C ‘pair’ 30o B a C

b = c a = b = c sin B sin C sin A sin B sin C A c b 25 cm 19 cm 30o B a C

b = c a = b = c sin B sin C sin A sin B sin C A 19 cm = 25 cm sin 30o sin C c b 25 cm 19 cm 30o B a C

b = c a = b = c sin B sin C sin A sin B sin C A 19 cm = 25 cm sin 30o sin C c b 19 (sin C) = sin 30o (25) 25 cm 19 cm 30o B a C

b = c a = b = c sin B sin C sin A sin B sin C A 19 cm = 25 cm sin 30o sin C c b 19 (sin C) = sin 30o (25) 25 cm 19 cm 19 (sin C) = (0.5)(25) 30o B a C

b = c a = b = c sin B sin C sin A sin B sin C A 19 cm = 25 cm sin 30o sin C c b 19 (sin C) = sin 30o (25) 25 cm 19 cm 19 (sin C) = (0.5)(25) 19 (sin C) = 12.5 30o B a C

b = c a = b = c sin B sin C sin A sin B sin C A 19 cm = 25 cm sin 30o sin C c b 19 (sin C) = sin 30o (25) 25 cm 19 cm 19 (sin C) = (0.5)(25) 19 (sin C) = 12.5 30o sin C = 0.658 B a C

b = c a = b = c sin B sin C sin A sin B sin C A 19 cm = 25 cm sin 30o sin C c b 19 (sin C) = sin 30o (25) 25 cm 19 cm 19 (sin C) = (0.5)(25) 19 (sin C) = 12.5 30o sin C = 0.658 B a C sin-1 (0.658) = C

b = c a = b = c sin B sin C sin A sin B sin C A 19 cm = 25 cm sin 30o sin C c b 19 (sin C) = sin 30o (25) 25 cm 19 cm 19 (sin C) = (0.5)(25) 19 (sin C) = 12.5 30o 41.1o sin C = 0.658 B a C sin-1 (0.658) = C C = 41.1o

Step 3. Find the last angle (A)
180o = A + B + C A c b 25 cm 19 cm 30o 41.1o B a C

180o = A + B + C A 180o = A + 30o o c b 25 cm 19 cm 30o 41.1o B a C

180o = A + B + C A 180o = A + 30o o c b 180o = A o 25 cm 19 cm 30o 41.1o B a C

180o = A + B + C A 180o = A + 30o o c b 180o = A o 25 cm 19 cm 180o – 71.1o = A 30o 41.1o B a C

180o = A + B + C A 108.9o 180o = A + 30o o c b 180o = A o 25 cm 19 cm 180o – 71.1o = A A = 108.9o 30o 41.1o B a C

Step 4. Find the last ‘pair’ (A)
a = b = c sin A sin B sin C A 108.9o c b 25 cm 19 cm 30o 41.1o B a C

a = b = c a = b sin A sin B sin C sin A sin B A 108.9o c b 25 cm 19 cm 30o 41.1o B a C

a = b = c a = b sin A sin B sin C sin A sin B A a = 19 cm 108.9o sin 108.9o sin 30o c b 25 cm 19 cm 30o 41.1o B a C

a = b = c a = b sin A sin B sin C sin A sin B A a = 19 cm 108.9o sin 108.9o sin 30o c b a = 19 cm 25 cm 19 cm 30o 41.1o B a C

a = b = c a = b sin A sin B sin C sin A sin B A a = 19 cm 108.9o sin 108.9o sin 30o c b a = 19 cm 25 cm 19 cm a (0.5) = (19) 30o 41.1o B a C

a = b = c a = b sin A sin B sin C sin A sin B A a = 19 cm 108.9o sin 108.9o sin 30o c b a = 19 cm 25 cm 19 cm a (0.5) = (19) a (0.5) = 17.97 30o 41.1o B a C

a = b = c a = b sin A sin B sin C sin A sin B A a = 19 cm 108.9o sin 108.9o sin 30o c b a = 19 cm 25 cm 19 cm a (0.5) = (19) a (0.5) = 17.97 30o 41.1o B a C a = 36 cm 36 cm

Done! A 108.9o c b 25 cm 19 cm 30o 41.1o B a C 36 cm

Steps to complete a triangle using Sine Law
Step 1. List the missing information, and what is needed Step 2. Find one ‘pair’, and use it to fill in another ‘pair’ Step 3. Find the last angle Step 4. Find the last ‘pair’

Uses the cos ratio Also uses ‘pairs’
Cos Law Uses the cos ratio Also uses ‘pairs’

pair you’re looking for
Looking for a Cos Law other two lengths a2 = b2 + c2 – 2bc(cosA) pair you’re looking for

Looking for b Cos Law other two lengths b2 = a2 + c2 – 2ac(cosB) pair you’re looking for

Looking for c Cos Law other two lengths c2 = b2 + a2 – 2ab(cosC) pair you’re looking for

3 variations of Cos Law a2 = b2 + c2 – 2bc(cosA)
b2 = a2 + c2 – 2ac(cosB) c2 = b2 + c2 – 2ab(cosC)

Find the length of b A c b 25 cm 30o B C a 36 cm

Find the length of b To use Cos Law, you have to know:
- One of the values of the pair you need (angle or length) - The two other lengths c b 25 cm 30o B C a 36 cm

What we know: mAB = c = 25 cm mBC = a = 36 cm B = 30o A c b 25 cm What we need: mAC = b = ? 30o B C a 36 cm

Step 2. Choose the variation of Cos Law that you need
a2 = b2 + c2 – 2bc(cosA) A b2 = a2 + c2 – 2ac(cosB) c c2 = b2 + c2 – 2ab(cosC) b 25 cm 30o B C a 36 cm

Step 3. Solve the equation
b2 = a2 + c2 – 2ac(cosB)

b2 = a2 + c2 – 2ac(cosB) b2 = – 2(36)(25)(cos30o)

b2 = a2 + c2 – 2ac(cosB) b2 = – 2(36)(25)(cos30o) b2 = – 2(36)(25)(0.866)

b2 = a2 + c2 – 2ac(cosB) b2 = – 2(36)(25)(cos30o) b2 = – 2(36)(25)(0.866) b2 = – 2(36)(25)(0.866)

b2 = a2 + c2 – 2ac(cosB) b2 = – 2(36)(25)(cos30o) b2 = – 2(36)(25)(0.866) b2 = – 2(36)(25)(0.866) b2 = –

b2 = a2 + c2 – 2ac(cosB) b2 = – 2(36)(25)(cos30o) b2 = – 2(36)(25)(0.866) b2 = – 2(36)(25)(0.866) b2 = – b2 =

b2 = a2 + c2 – 2ac(cosB) b2 = – 2(36)(25)(cos30o) b2 = – 2(36)(25)(0.866) b2 = – 2(36)(25)(0.866) b2 = – b2 = b = 19 cm

Done! A c b 25 cm 19 cm 30o B a C 36 cm

Steps to complete triangles using Cos Law
Step 1. List the missing information, and what is needed Step 2. Choose the variation of Cos Law that you need Step 3. Solve the equation

How do you know which to use?
Use Sin Law if: Use Cos Law if: Given 2 sides, 1 angle opposite one of the sides Given 2 angles, 1 side opposite one of the angles Given 3 sides Given 1 angle, 2 sides adjacent to that angle a a c A b b B c a C C b

Summary of trigonometry
Right-angled triangle Non-right angled triangle If you only have sides --Pythagoras If you have sides and angles SOH CAH TOA If you have a pair (a + A) Sin Law If you have to fill in a pair (looking for angle or side) Cos Law a2 = b2 + c2 – 2ac(cos A)

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