Chapter 6: Trigonometry6.2: Trigonometric Applications

Essential Question:What does SOHCAHTOA mean and how

is it used to solve a right triangle problem?

6.2: Trigonometric Applications

• Old stuff will be used in this section– Triangle Sum Theorem• The sum of the measures of the angles in a triangle is

180°

– Pythagorean Theorem• In a right triangle with legs a and b and hypotenuse c,

a2 + b2 = c2

6.2: Trigonometric Applications

• Finding a side of a Triangle– Find side x in the right triangle below

– In this figure, we’re given:An angle (65°)The Hypotenuse (8)A side Adjacent to 65° (x)

The sides we’re using are A and Husing SOH-CAH-TOA

means we use the cosine function65°

8

x

8cos65 8 8

3.3809

x

x

6.2: Trigonometric Applications

Finding an Angle of a Triangle› Find the measure of the angle θ in the triangle below

In this triangle, we’re given all threeside lengths, so we can use any ofthe trigonometric ratios to solve.

› SOH sin θ = 3/5 → sin-1(3/5) = 36.8699°

› CAH cos θ = 4/5 → cos-1(4/5) = 36.8699°

› TOA tan θ = 3/4 → tan-1(3/4) = 36.8699°

› All ratios give us the same answer: 36.8699°

θ

4

5

3

6.2: Trigonometric Applications

• Solving a Right Triangle– Solve the right triangle below

– The Triangle Sum Theorem helps find θ

75° + θ + 90° = 180°θ = 15°

We can use the hypotenuse (17) and the

75° angle to find sides a and b75°

17

b

a

θ

17

17

sin 75 17 17

16.42

cos75 17 17

4.40

a

b

a

b

6.2: Trigonometric Applications

Solving a Right Triangle› Solve the right triangle below

The Pythagorean Theorem helps find aa2 + 62 = 122

a2 = 108a =

We can find β by using the cosine functioncos β = 6/12cos β = 1/2cos-1(1/2) = β60° = β

We can either find θ by using the sin function

or by using The Triangle Sum Theorem θ = 30°

β

12

6

a

θ108 6 3

6.2: Trigonometric Applications

• Assignment– Page 429• Problems 1 – 35, odd problems• Questions where you’re told to not use a calculator can

be solved using the chart you copied on Tuesday.

Chapter 6: Trigonometry6.2: Trigonometric Applications

Day 2Essential Question:

What does SOHCAHTOA mean and how is it used to solve a right triangle

problem?

6.2: Trigonometric Applications

Applications› A straight road leads from an ocean beach at a constant upward

angle of 3°. How high above sea level is the road at a point 1 mile from the beach? Answer

If one is not drawn, DRAW A DIAGRAM.

Looking for the side on the right of the triangle, which is the side opposite of 3°

sin 3° = x/52805280 • sin 3° = x276.33 ft = x

6.2: Trigonometric Applications

• Applications– According to the safety sticker on a 20-foot ladder, the distance from

the bottom of the ladder to the base of the wall on which it leans should be one-fourth of the length of the ladder: 5 feet.• How high up the wall will the ladder reach• If the ladder is in this position, what angle does it make with the ground?

• Draw a diagram

6.2: Trigonometric Applications

Applications› The wall height can be found

using the Pythagorean Theorem 52 + h2 = 202

h2 = 400 – 25 h2 = 375 h = (375)½ ≈ 19.36 ft

› We’re given the side adjacent to θ and the hypotenuse, meaning we need to use cosine cos θ = 5/20 θ = cos-1(5/20) ≈ 75.5°

6.2: Trigonometric Applications

Angles of Elevation and Depression

› Both create right angles from an endpoint. Angles of elevation look up; angles of depression look down.

Angle of elevation

Angle of depression

6.2: Trigonometric Applications

Elevation/Depression› A flagpole casts a 60-foot shadow when the angle of elevation of the

sun is 35°. Find the height of the flagpole. Draw a diagram You’re given a 35° angle

You’re given the side adjacent You’re looking for the side opposite

You’re using tangent tan 35° = x / 60 60 • tan 35° = x 42.012 ≈ x

6.2: Trigonometric Applications

• Elevation/Depression (#3: both)– A person on the edge of a canal observes a lamp post on the other

side with an angle of elevation of 12° to the top of the lamp post and an angle of depression of 7° to the bottom of the lamp post from eye level. The person’s eye level is 152 cm.• Find the width of the canal.• Find the height of the lamp post.• Draw a diagram

12°

7°152 cm

152 cm

canal

top half of lamp post

6.2: Trigonometric Applications Elevation/Depression (#3: both)

› The canal is adjacentto the 7° angle

› You’re given 152 cm, which is opposite 7° Use tangent tan 7° = 152 / x x = 152 / tan 7° x = 1237.94 cm

› Use the canal measurement to find the top half of the lamp post… again using tangent. tan 12° = y / 1237.94 1237.94 • tan 12° = y 263.13 cm = y

› So the height of the lamp post is 263.13 + 152 = 415.13 cm

12°

7°152 cm

canal (x)

top half of lamp post (y)

6.2: Trigonometric Applications

Assignment› Page 431

Problems 37 – 49, odd problems

› Quiz tomorrow1)DMS/decimal conversion2)Finding 6 trig ratios3)Solving right triangles4)A word problem or two