Big Ideas Math Geometry, 2014
BI
Big Ideas Math Geometry, 2014 View details
7. Law of Sines and Law of Cosines
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Exercise 51 Page 516

Practice makes perfect
a Let's consider the triangle ABC shown below.

Next, let's draw the altitude from vertex B to AC. We will label the altitude as h.

Notice that both △ ADB and △ CDB are right triangles, so we can apply the trigonometric ratios to write the two equations below. ccc sin A = h/c & & sin C = h/a ⇕ & & ⇕ h = csin A & & h = asin C Recall that the area of a triangle is one half the base multiplied by the height. A = 1/2 b * h By using each of the heights written above, we can obtain two of three required formulas. A = 1/2 b * h ↙ ↘ A = 1/2 b csin A A = 1/2 b asin C To obtain the remaining formula, we have to draw any of the other two altitudes. For example, let's draw the altitude from A.

Next, let's write the sine ratio of ∠ B. ccc sin B = h_1/c ⇒ h_1 = csin B Finally, we can find the area of △ ABC by using a as the base and h_1 as the height. A = 1/2 a * h_1 ⇒ A = 1/2 a csin B

b Each of the three expressions in Part A represent the area of the same triangle. Therefore, we can equate them and obtain the equation below.

1/2 b csin A = 1/2 a csin B = 1/2 a bsin C

c In Part B we set the following equation.
1/2 b csin A = 1/2 a csin B = 1/2 a bsin C

Let's rewrite the equation above. First, let's multiply each side by 2 to get rid of the fractions. b csin A = a csin B = a bsin C Next, let's divide each side by a b c and cancel the common factors. b* c* sin A/a* b* c = a* c * sin B/a* b* c = a* b* sin C/a* b* c [0.1cm] ⇕ [0.1cm] sin A/a = sin B/b = sin C/c We have now obtained the equations given by the Law of Sines.