Bell Work Find the measure of angle A Find x. A 7” 9” 30° 10” x

Bell work #1 Solution A 7” 9” hypotenuse opposite

Bell Work #2 30° 10” x

Introduction Given a right triangle, you should feel comfortable using the three basic trig functions( sin, cos, tan) to determine additional information about the triangle.

Law of Sines Apply Law of Sines to problem-solving situations Find the area of triangle.

The Law of Sines Law of Sines Alternative forms are sometimes convenient to use: Law of Sines In any triangle ABC, with sides a, b, and c,

Data Required for Solving Oblique Triangles Case 1 One side and two angles known: SAA or ASA Case 2 Two sides and one angle not included between the sides known: SSA This case may lead to more than one solution. Case 3 Two sides and one angle included between the sides known: SAS Case 4 Three sides are known: SSS

Law of sines Case 1 One side and two angles known: SAA or ASA Case 2 Two sides and one angle not included between the sides known: SSA This case may lead to more than one solution.

Example 1

Law of Sines Example 1 Given the diagram below, determine the length of side x. 47° 63° 5.45” x

Law of Sines Example 1 What do we know about this problem? First of all, it is an oblique triangle. Second, we note that two angles are known, and one of the sides opposite. 47° 63° 5.45” x

Law of Sines Example 1 That’s enough info to verify that using the Law of Sines will allow us to determine the length of x. 47° 63° 5.45” x

Law of Sines Example 1 To solve, set up a proportion. Remember that the sides are proportional to the sines of the opposite angles. 47° 63° 5.45” x

Law of Sines Example 1 Start by pairing the 63° angle and the 5.45” side together since they are opposite one another. 47° 63° 5.45” x

Law of Sines Example 1 The unknown side x is opposite the 47° angle. Pair these up to complete the proportion. 47° 63° 5.45” x

Law of Sines Example 1 Solve the proportion by cross-multiplying. 63° Multiply on this diagonal first. 5.45 x sin47° = 3.99 47° 63° 5.45” x

Law of Sines Example 1 Solve the proportion by cross-multiplying. 63° Next, divide 3.99 by sin63° 3.99 ÷ sin63° = 4.47 (this is the length of x) 47° 63° 5.45” x

Law of Sines Example 1 By using the Law of Sines, we know the length of side x is 4.47 inches. 47° 63° 5.45” 4.47”

Example 2

Law of Sines Example 2 Given the diagram below, determine the length of side x. 42° 85° 65.85 mm x

Law of Sines Example 2 Before you jump in, be sure you know what you are dealing with. You are working with an oblique triangle... …and you know two angles and a side opposite one of those angles. 42° 85° 65.85 mm x

Law of Sines Example 2 That means using the Law of Sines will allow you to solve for x. 42° 85° 65.85 mm x

Law of Sines Example 2 Set-up a proportion, starting with the 65.85 mm side and the 85° angle since they are opposite one another. 42° 85° 65.85 mm x

Law of Sines Example 2 Solve the proportion. 42° 85° 65.85 mm x Multiply on this diagonal first. 65.85 x sin42° = 44.1 42° 85° 65.85 mm x

Law of Sines Example 2 Solve the proportion. 42° 85° 65.85 mm x Multiply on this diagonal first. 65.85 x sin42° = 44.1 42° 85° 65.85 mm x

Law of Sines Example 2 Solve the proportion. 42° 85° 65.85 mm x Next, divide 44.1 by sin85° 44.1 ÷ sin85° = 44.2 (this is the length of x) 42° 85° 65.85 mm x

Law of Sines Example 2 Using the Law of Sines on this problem gives you an answer of 44.2 mm. 85° 42° 44.2 mm 65.85 mm

Example 3

Law of Sines Example 3 Try this one on your own. Set-up a proportion and solve for x. 9.25 cm 88° x 57°

Law of Sines Example 3 How did it turn out? x = 11.02 cm 9.25 cm 88° x 57°

Example 4

Law of Sines Recall that the other scenario where you can use the Law of Sines is when you know the lengths of two sides and the size of an angle opposite on of those sides. 42° 85.5 mm 70 mm

Law of Sines Example 4 Given the diagram below, determine the size of angle A. 42° 85.5 mm 70 mm A

Law of Sines Example 4 Once again, set-up a proportion. Start by pairing-up the 70 mm side and the 42° angle. 42° 85.5 mm 70 mm A

Law of Sines Example 4 Complete the proportion by putting the 85.5 mm side and angle A together. 42° 85.5 mm 70 mm A

Law of Sines Example 4 This proportion will be a little more difficult to solve. The steps are shown below: Cross-multiply on this diagonal...

Law of Sines Example 4 This proportion will be a little more difficult to solve. The steps are shown below: Cross-multiply on this diagonal...

Law of Sines Example 4 This proportion will be a little more difficult to solve. The steps are shown below: Cross-multiply on this diagonal...

Law of Sines Example 4 This proportion will be a little more difficult to solve. The steps are shown below: Divide both sides by 70.

Law of Sines Example 4 This proportion will be a little more difficult to solve. The steps are shown below: Evaluate the left side of the equation.

Law of Sines Example 4 This proportion will be a little more difficult to solve. The steps are shown below: Evaluate the left side of the equation.

Law of Sines Example 4 This proportion will be a little more difficult to solve. The steps are shown below: Type 0.8173 into your calculator, press the 2nd function key, then press the sin key.

Law of Sines Example 4 So the size of angle A is 54.8°. 42° 85.5 mm

Example 5

Law of Sines Example 5 Given the diagram below, determine the size of angle A. 55° 4.9” 4.2” A

Law of Sines Example 5 Once again, set-up a proportion. Start by pairing-up the 4.2” side and the 55° angle. 55° 4.9” 4.2” A

Law of Sines Example 5 Complete the proportion by putting the 4.9” side and angle A together. 55° 4.9” 4.2” A

Law of Sines Example 5 Follow the steps shown to solve the proportion: Cross-multiply on this diagonal...

Law of Sines Example 5 Follow the steps shown to solve the proportion: …then multiply on this diagonal...

Law of Sines Example 5 Follow the steps shown to solve the proportion: Divide both sides by 4.2.

Law of Sines Example 5 Follow the steps shown to solve the proportion: Evaluate the left side of the equation.

Law of Sines Example 5 Follow the steps shown to solve the proportion: Type 0.9557 into your calculator, press the 2nd function key, then press the sin key.

Law of Sines Example 5 You have just determined that angle A is 72.9°. 4.2” 72.9° 4.9” 55°

Example 6

Law of Sines Example 6 Try this one on your own. Set-up a proportion and solve for angle A. Then click to see the answer. 10.8” A 12.25” 60°

Law of Sines Example 6 The set-up and answer are shown below: 10.8” A 12.25” 60°

Area of Triangle

We are all familiar with the formula for the area of a triangle, where b stands for the base and h stands for the height drawn to that base.

In the triangle at the right, the area could be expressed as:

Now, let's be a bit more creative and look at the diagram again Now, let's be a bit more creative and look at the diagram again.  By using the right triangle on the left side of the diagram, and our knowledge of trigonometry, we can state that: This tells us that the height, h, can be expressed as bsinC.

If we substitute this new expression for the height, we can write the area formula as:  

We have just discovered that the area of a triangle can be expressed using the lengths of two sides and the sine of the included angle.  This is often referred to as the SAS Formula for the area of a triangle.  The "letters" in the formula may change from problem to problem, so try to remember the pattern "two sides and the sine of the included angle".

"Wow! A trig area formula for triangles!!!" We no longer have to rely on a problem supplying us with the length of the altitude (height) of the triangle.  If we know two sides and the included angle, we are in business.

Example 1: Given the triangle at the right, find its area.  Express the area rounded to three decimal places. Be careful!!!  When using your graphing calculator, be sure that you are in DEGREE Mode, or that you are  using the degree symbol.

Given the parallelogram shown at the right, find its EXACT area. If we are looking for an EXACT answer, we do NOT want to round our value for sin 60º.  We need to remember that the sin 60º (from our 30º- 60º- 90º reference triangle is_______________ . Now, the diagonal of a parallelogram divides the parallelogram into two congruent triangles.  So the total area of the parallelogram will be square units.

Given the parallelogram shown at the right, find its EXACT area. If we are looking for an EXACT answer, we do NOT want to round our value for sin 60º.  We need to remember that the sin 60º (from our 30º- 60º- 90º reference triangle is .

Given the parallelogram shown at the right, find its EXACT area.

Now, the diagonal of a parallelogram divides the parallelogram into two congruent triangles.  So the total area of the parallelogram will be square units

End of Lesson Let’s Practice.