a Draw the altitude from B to AC and find the sine ratio of ∠A and ∠C. Substitute the obtained expressions into the formula of the area. Repeat the process with another altitude.
B
b What does each formula represent?
C
c Multiply by a respective number to get rid of the fractions. Then, divide by the product of the three side lengths and cancel the common factors.
A
a See solution.
B
b The three formulas represent the area of â–³ ABC.
C
c See solution.
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 â–³ ADB and â–³ 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 â–³ 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 CLet'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.
