Unit 5 – Trig Laws(Pre-Calculus)

Unit 6 Warm-

Ups

Day 1

The Law of Sines (1 solution)

Day 2

The Law of Sines

(ambiguous case)

Day 3

The Law of Cosines wskt

Day 4

The Laws of Sine and

Cosine and Area wkst

Day 5

The Laws of Sine and

Cosine Partner Challenge

Day 6

The Laws of Sine and

Cosine Word Problems wkst

Day 7

More Law Applications

wkst

Review Wkst(optional)

Quick Check #1

Triangle Area

Quick Check #2

Law of Sines

Quick Check #3

Law of Cosines

Unit Circle Qz #3

Unit Circle Qz #4

Name _______________Score _________

Quotes written on back YES NO

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Unit 5 – Trig LawsDay 1 * Finding Area of Oblique

Triangles warm-up* New Area Formulas* Solving Oblique Triangles using the Law of Sines Notes

* Day 1: Law of Sines (1 solution) wkst

Day 2 * Law of Sines – ambiguous case notes and practice * Day 2: Law of Sines (ambig.

Case) wkst

Day 3 * Quick Check #1 – Area of Triangles* Law of Sines Review* Introduce the Law of Cosines and Herons Formula

* Day 3: Law of Cosines wkst

Day 4 * Quick Check #2: Law of Sines* Law of Cosines Practice

* Day 4: The Laws of Sine / Cosine wkst

Day 5 * Laws of Sine/Cos Partner Challenge* Begin working on Applications involving Trig Laws

Day 6 * Quick Check #3: Law of Cosine* More Application work

* Day 6 – Laws of Sin/Cos Word Problems wkst

Day 7 * Partner EC – Lot Problem* Review for Test * Day 7 – More Application

WorkDay 8 TEST DAY

Finding AREA using Trig

Find the AREA of the following triangles using whatever method you like. (keep in mind the area of a triangle is 1/2

Base*Height)

a) b)

Now that you have tried these out, here’s an easier way to find the area – using a little trigonometry.

* If you know two sides of any triangle and the angle measure that is between the two sides you can use this formula to find the area.

Area = ½ (side)(side) (sine of angle in between sides)

or

Area = ½ ab sin C, or ½ bc sin A, or

½ ac sin B

Example: Area = ½(28m)(35m)(sin 1240) ≈ 406.23 m2

Use this formula to find the area of the first two triangles given above.

a) b)Here’s another AREA formula (this one does not use trig, but it still can be very useful – it’s called Heron’s Formula).

Example: s = (12+23+30)/2 = 32.5Area = √s (s−a )(s−b)(s−c )

s = (a+b+c)/2 Area= 125.8 m2

√32.5(32.5−12)(32.5−23)(32.5−30)

Find the area of each triangle using Heron’s Formula:

7) 8) 9)

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The “LAW of SINES” or

These equations are used to solve oblique triangles. You must be given the following to use this law.

1) Two angles and any side (AAS or ASA) 2) Two sides and an angle opposite one of them (SSA)

(This second case is an ambiguous case)

* To label a triangle, the vertices are labeled with capital letters and the sides are labeled with lower case letters. Side 'a' is opposite angle 'A'. This follows for the other sides and angles also.

Ex. 1 Solve the triangle: (this means to find all the missing side lengths and angles)

A) Set up a proportion using the given information and one missing piece:

B) Use cross products to solve the D) Use the angle you found in proportion: (c) to find the final side:

C) You can find the missing angle by subtracting the given angles from 180o.

Ex 2: Solve the Triangle.

Day 2: The AMBIGUOUS CASE (SSA)

* Rule for possible ambiguous situations:

At the key step, find your angle using the inverse sine (sin-1) function on your calculator. After you find that angle, subtract it from 180o to find the other possible angle

with the same sine value. Add this angle with the other angle you know. If these two angles have a sum less

than 180o, then you will have 2 possible solutions for your triangle.

Ex. 4:

Ex. 5:

Ex. 6:

Day 3: Law of Sines Review

1)

2)

3)

Day 4 – Law of Cosines

If you look at the two triangles above, you will notice that neither of them would be able to be solved using the Law of Sines (you need at least one angle and its opposite side to use this). In order to solve these triangles for the missing sides and angles you will need to use the Law of Cosines. This formula is a little longer than the Law of Sines, but just as easy to use.

Law of Cosinesa2 =

b2 =

c2 =

** To help in finding the angle measures, you can solve the given equations for cos A, cos B, or cos C. Then the Law of Cosines looks like this **

Ex. 1:

* In this next example, you are given all the side lengths of the figure. When this happens, YOU MUST FIND THE LARGEST ANGLE FIRST – THIS IS THE ANGLE ACROSS FROM THE LONGEST SIDE. If you fail to do this, your answer might be incorrect.

Ex. 2:

Practice:

Step 1: Use the Law of Cosines to find side ‘a.’ a

a2 = b2 + c2 – 2bc (cos A)

a2 = ___________________________________________________

a = _________________ (round to two decimal places)

Step 2: Use the Law of Sines to find angle B.

Angle B = ____________

Step 3: Since you know angle A and B now, find angle C by subtracting A and B from 1800.

Angle C = _______________

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Step 1: Find the largest angle, B.

cos B = = ______________ (decimal value)

cos-1( ) = _______________ =

Step 2: Use the Law of Sines to find angle A.

= = _______________

Step 3: Find the value of angle C by subtracting A and B from 1800.

= _____________

Day 5: ApplicationsEx. 1: David and Angela are standing at the seashore 3 miles apart. The coastline is a straight line between them. Both can see the same ship in the

water. The angle between the coastline and the line between the ship and David is 55 degrees. The angle between the coastline and the line between the ship and Angela is 45 degrees. How far is the ship from David?

Ex. 2: Airplane A is flying directly toward the airport which is 35 miles away. The pilot notices airplane B 50 degrees to her right. Airplane B is also flying directly toward the airport. The pilot of airplane B calculates that airplane A is 63 degrees to his left. Based on that information, how far is airplane B from the airport?

Ex. 3: A plane leaves OHare Airport and travels due east at 425 mi/hr. Another plane leaves 20 minutes later and travels 27º east of south at the rate of 475 mi/h. To the nearest ten miles, how far apart are they 1 hour after the second plane leaves.

Ex. 4: After the hurricane, the small tree in my neighbor’s yard was leaning. To keep it from falling, we nailed a 8-foot strap into the ground 5 feet from the base of the tree. We attached the strap to the tree 4 feet up the slanted tree. How far from vertical was the tree leaning?