CHAPTER 3

Acute Triangle Trigonometry

Chapter 3: Acute Triangle Trigonometry

3.1 – Exploring Side-Angle Relationships in Acute Triangles

Trigonometry

What are the three trigonometric functions?

Sine (sin), Cosine (cos), and Tangent (tan)

Remember:

SOH CAHTOA

Hypote

nuse

Opposit

esin

Ad

jace

nt

Opposite

tan

Hypote

nuse

Ad

jace

nt

cos

Example

Solve for height, h:sin(46.5) = h/51.2cm (sin=opp/hyp) h = 51.2 × sin(46.5) h = 37.1 cm

Or:sin(83.3) = h/37.3cm h = 37.3 × sin(83.3) h = 37.1 cm

So:37.3 × sin(83.3) = 51.2 × sin(46.5)

Or:

• What’s the relationship between 37.3 cm and 46.5°?

• How about 51.2 cm and 83.3°?

• What would have happened if we had put the h going from vertex B down to its opposite side? What could we say about 39.7 cm and 50.4°?

The Sine Law

Law of Sines:

OR

PG. 131, # 1-5

Independent Practice

Chapter 3: Acute Triangle

3.2 – Proving and Applying the Sine Law

Proof:

h2

Draw a triangle like the last, with another perpendicular line from b to vertex B, called h2.

sin(C) = h/bsin(B) = h/c

sin(A) = h2/csin(C) = h2/a

h = b x sin(C) = c x sin(B) b = c . sin(B) sin(C)

h2 = c x sin(A) = a x sin(C) c = a . sin(C) sin(A)

Last time we learned the Sine Law:

Remember!sin = opp/hyp

Example

A triangle has angles measuring 80° and 55°. The side opposite the 80° angle is 12.0 m in length. Determine the length of the side opposite the 55°angle to the nearest tenth of a metre.

Draw a diagram: What will the third angle be? What’s the total sum of angles in a triangle?

Sine Law:

Which side are we looking for?

To the nearest tenth of a metre, the side opposite the 55° angle is 10.0 m long.

Example

Toby uses chains attached to hooks on the ceiling and a winch to lift engines at his father’s garage. The chains, the winch, and the ceiling are arranged as shown. Toby solved the triangle using the sine law to determine the angle that each chain makes with the ceiling to the nearest degree. He claims that θ=40° and α=54°. Is he correct? Explain, and make any necessary corrections.

So, if θ=51°, how can we tell what α is? Remember, the sum of the angles must be 180°.

θ+α+86° = 180°51°+α+86° = 180°α=180°-51°-86°=43°

So, Toby was incorrect, in actuality θ=51° and α=43°.

Example

The captain of a small boat is delivering supplies to two lighthouses, as shown. His compass indicates that the lighthouse to his left is located at N30°W and the lighthouse to his right is located at N50°E. Determine the compass direction he must follow when he leaves lighthouse B lighthouse A.

The angle B is 47.612°, but we want the directionhe’d go from lighthouse B.

We need to draw a line pointing East from vertex B.Since the new line goes West to East, it must be perpendicular to the line going South to North… so the angle is 90°.

If we know that one angle in ΔBCO is 90° and another is 50°, how can we figure out the third angle?

O

∠CBO = 180° - 50° - 90° = 40°

∠NBO = 47.6° - 40° = 7.6°

To find the direction on the compass, we must subtract our answer from 90°. So, he will head N82°W from lighthouse B.

PG. 138-141, # 2-15, 18

Independent Practice

Chapter 3: Acute Triangle Trigonometry

3.3 – Proving and Applying the Cosine Law

Cosine Law

In any acute triangle,a2 = b2 + c2 – 2bc

cosAb2 = a2 + c2 – 2ac

cosBc2 = a2 + b2 – 2ab

cosC

Proof of the Cosine Law

In any acute triangle,a2 = b2 + c2 – 2bc

cosAb2 = a2 + c2 – 2ac

cosBc2 = a2 + b2 – 2ab

cosCProof:

h2 = c2 – x2

h2 = b2 – y2

c2 – x2 = b2 – y2

c2 = x2 + b2 – y2

c2 = (a – y)2 + b2 – y2 (since x = a – y)

c2 = a2 – 2ay + y2 + b2 – y2

c2 = a2 + b2 – 2ay

cosC = y/b (cos = adj/hyp)b cosC = y

c2 = a2 + b2 – 2ayc2 = a2 + b2 – 2ab cosC

The other parts can be proven similarly.

Example

Determine the length of CB to the nearest metre.

Why can’t you use the Sine Law?

a2 = b2 + c2 – 2bc cosAa2 = 322 + 402 – 2(32)(40)cos(58°)

a2 = 1024 + 1600 – 2560cos(58°)a2 = 2624 – 2560cos(58°)a2 = 1267.406a = 35.600…

CB is 36 m.

Example

A three-pointed star is made up of an equilateral triangle and three congruent isosceles triangles. Determine the length of each side of the equilateral triangle in this three-pointed star. Round the length to the nearest centimetre.

First, label the diagram!x2 = z2 + y2 – 2zy cosXx2 = 602 + 602 – 2(60)(60)cos(20°)x2 = 3600 + 3600 – 6765.786…

x2 = 434.213x = 20.837…

Each side of the equilateral triangle has a length of 21 cm.

PG. 150 – 153, #1, 4, 5, 7, 8, 10, 11, 13, 15.

Independent Practice

Chapter 3: Acute Triangle Trigonometry

3.4 – Solving Problems Using Acute Triangles

Example

Two security cameras in a museum must be adjusted to monitor a new display of fossils. The cameras are mounted 6 m above the floor, directly across from each other on opposite walls. The walls are 12 m apart. The fossils are displayed in cases made of wood and glass. The top of the display is 1.5 m above the floor. The distance from the camera on the left to the centre of the top of the display is 4.8 m. Both cameras must aim at the centre of the top of the display. Determine the angles of depression, to the nearest degree, for each camera.

Draw a diagram:sinθ=4.5/4.8 θ=sin-1(4.5/4.8)θ=69.635…°

Example continued…

z2 = d2 + b2 – 2db cosAz2 = 122 + 4.82 – 2(12)(4.8) cos(69.635)z2 = 126.952…z = 11.267…

sinα= 4.5/11.267α= sin-1(4.5/11.267)α= 23.539°

Camera A must be adjusted to an angle of 70° and camera B must be adjusted to an angle of 24°.

Example

Brendan and Diana plan to climb the cliff at Dry Island Buffalo Jump, Alberta. They need to know the height of the climb before they start. Brendan stands at point B, as shown in the diagram. He uses a clinometer to determine ∠ABC, the angle of elevation to the top of the cliff. Then he estimates ∠CBD, the angle between the base of the cliff, himself, and Diana, who is standing at point D. Diana estimates ∠CDB, the angle between the base of the cliff, herself, and Brendan. Determine the height of the cliff to the nearest metre.In ΔDBC,

∠BCD = 180° - 60° - 50°∠BCD = 70°

Example continued…

The height of the cliff is 196 m.

PG. 161 – 164, #1-14, 16.

Independent Practice

Activity

Cut out the seven similar triangles from the handout, and use all seven of them to form a single square. Show me once you succeed.

If the hypotenuse of the greatest triangle is 10 units long, what is the area of the square?