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Alternate Formula for the Area of a Triangle

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Sine to find the Area of a Triangle

You have previously learned that the area of a triangle is A=\frac{1}{2}bh  where  b is a base of the triangle and  h is the corresponding height. Why is it helpful to have another formula for calculating the area of a triangle?

Watch This

http://www.youtube.com/watch?v=WEOVjRMh_vI James Sousa: Determine the Area of a Triangle Using the Sine Function

Guidance

One way to find the area of a triangle is by calculating \frac{1}{2}bh . This formula works when you know or can determine the base and the height of a triangle. What if you wanted to find the area of the following triangle:

The two given sides are not the base and the height. In the examples you will derive a formula for calculating the area of a triangle given this type of information.

Example A

Below, the altitude of \Delta ABC  from  A has been drawn. Write an equation that relates  h, 34^\circ and 7.

Solution: Notice that two right triangles have been formed. This means you can use the trigonometric ratios to relate the sides.   h is opposite the 34^\circ  angle and 7 is the length of the hypotenuse of the right triangle. This is the sine relationship.

\sin 34^\circ=\frac{h}{7}

Example B

Solve the equation from Example A for h . Then, use A=\frac{1}{2}bh  to find the area of the triangle using your value for h . Can you generalize the formula?

Solution: If  \sin 34^\circ=\frac{h}{7} then h=7 \sin 34^\circ . For the height that has been drawn, \overline{BC}  is the base.

A &= \frac{1}{2}bh\\A &= \frac{1}{2} (12)(7 \sin 34^\circ)\\A &= \frac{1}{2} (12)(7) (\sin 34^\circ)\\A & \approx 23.45 \ un^2

Notice the calculation: A =\frac{1}{2} (12) (7) (\sin 34^\circ) . You found half the product of the two sides and the sine of the included angle. Consider the general triangle below:

If you know \angle B , the area will be Area = \frac{1}{2} (a)(c) \sin B . If you know \angle C , the area will be Area=\frac{1}{2} (a)(b) \sin C .

Example C

Previously, you have only used the sine ratio on acute angles. Can you find the area of a triangle when given an obtuse angle? Consider the triangle below. Find an equation to determine the value of h . Then, find the area of the triangle.

Solution: First find the measure of the exterior angle of the triangle at vertex B . Remember that the interior and exterior angles at vertex  B must sum to 180^\circ .

Now, consider the right triangle with \overline{AB}  as the hypotenuse.  h is the side opposite the 55^\circ  angle and 7 is the hypotenuse. Once again, this is the sine relationship.

\sin 55^\circ &= \frac{h}{7}\\\rightarrow h &= 7 \sin 55^\circ

For the height that has been drawn, \overline{BC}  is the base.

A &= \frac{1}{2} bh\\A &= \frac{1}{2} (12)(7 \sin 55^\circ)\\A &= \frac{1}{2} (12) (7)(\sin 55^\circ)\\A & \approx 34.4 \ un^2

In general, if your given angle is obtuse, you can use the angle supplementary to your given angle in your area calculation.

Concept Problem Revisited

You have previously learned that the area of a triangle is A=\frac{1}{2}bh  where  b is a base of the triangle and  h is the corresponding height. Why is it helpful to have another formula for calculating the area of a triangle?

Often you only know the sides and angles of a triangle and not the height.  A=\frac{1}{2} (a)(b) \sin C allows you to quickly find the area of acute or obtuse triangles when given two sides and an included angle.

Vocabulary

The sine (sin) of an angle within a right triangle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

The trigonometric ratios are sine, cosine, and tangent. 

Trigonometry is the study of triangles.

\theta , or “theta”, is a Greek letter. In geometry, it is often used as a variable to represent an unknown angle measure.

Guided Practice

Find the area of each triangle.

1.

2.

The sine ratio is actually a function that takes any angle measure as an input (not just angles between 0^\circ  and 90^\circ ). One property of the sine function is that the sine of supplementary angles will always be equal. Therefore, when finding the area of a triangle given an obtuse angle, you can use the obtuse angle in your calculation instead of the acute supplementary angle and your answer will be the same. Note: You will study the sine function in much more detail in future courses!

3. Find the area of the triangle from #2 using the obtuse angle in your calculations. What do you notice?

Answers:

1. The 65^\circ  angle is the included angle to the two given sides of length 7 and 10. Use the new area formula. 

A &= \frac{1}{2} (7)(10) \sin 65^\circ\\A & \approx 31.72 \ un^2

2. Notice that the given angle is obtuse. Draw the altitude from vertex  C so that it intersects the extension of  \overline{AB} at point D . The exterior angle at \angle A  is 30^\circ .

Use the new area formula with the 30^\circ  angle.

A &= \frac{1}{2} (9)(15) \sin 30^\circ\\A &= 33.75 \ un^2

3. Use the area formula as in #2. Instead of using 30^\circ , use 150^\circ .

A &= \frac{1}{2} (9)(15) \sin 150^\circ\\A &= 33.75 \ un^2

The result is the same.  \sin 30^\circ=0.5 and \sin 150^\circ=0.5 .

This means that regardless of whether the given angle is acute or obtuse, you can always find the area of a triangle by finding the half the product of two sides and the sine of their included angle:

For all triangles:  A=\frac{1}{2} (a)(b) \sin C

Practice

Find the area of each triangle.

1.

2.

3.

4.

5. Explain why the following triangle has the same area as the triangle in #4.

Find the area of each triangle.

6.

7.

8.

9. Use your calculator to find \sin 90^\circ .

10. Find the area of the triangle below in two ways. First, use the formula A=\frac{1}{2} bh . Then, use the new formula from this concept. What do you notice?

11. Find the area of the parallelogram:

12. Use your work from #11 to help you to describe a general method for calculating the area of a parallelogram given its sides and angles.

13. The area of the triangle below is 78 \ un^2 . Find the measure of \theta  rounded to the nearest degree.

14. Use the triangle below to explain where the area formula  A=\frac{1}{2} ab \sin C comes from.

15. Why does the area formula A=\frac{1}{2} ab \sin C  work even if \angle C  is obtuse?

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