← Back to course
Precalculus 9-12 / Week 08 / Thursday
4/6
Week 08 · Laws of Sines and Cosines

Thursday

Why the laws are true
// The buoy sighted from shore
⏱ about 20 min

Thursday: Why the Laws Are True

"Yesterday we used the laws," Wren says, chalking a big triangle on the dock. "Today we say why they hold."

He drops a dashed line from C straight down to side c. "One height, h. It cuts the triangle into two right ones."

"In the left one, sin A = h over b," Comet says. "In the right one, sin B = h over a."

Nova glows the two equations side by side. "Would you like a hint? Both equal h. Set them equal."

"b sin A = a sin B," Wren writes. "Divide by sin A and sin B. There is the Law of Sines."

"And area is half of c times h," Comet adds. "Replace h with b sin A. Half of c b sin A."

"Pythagoras on the same two right triangles gets us the Law of Cosines," Wren says. "Three proofs, one height."

Derivation one: area = ½ab sin C

  1. Draw triangle ABC. Drop a height h from B straight down to side b (the side AC).
  2. Area of any triangle is ½ × base × height, so area = ½ × b × h.
  3. The height h is the side opposite angle C in a right triangle whose longest side is a.
  4. So sin C = h/a, which means h = a sin C.
  5. Replace h: area = ½ × b × a sin C = ½ab sin C.

Check with the ropes: ½ × 8 × 11 × sin 52°. That is ½ × 8 × 11 × 0.788 ≈ 34.7 square meters.

Derivation two: the Law of Sines, read first

  1. Drop a height h from C to side c. It makes two right triangles that share h.
  2. In the right triangle at A: sin A = h/b, so h = b sin A.
  3. In the right triangle at B: sin B = h/a, so h = a sin B.
  4. Both equal h, so b sin A = a sin B.
  5. Divide both sides by sin A sin B: a/sin A = b/sin B.
  6. Drop a height from A instead and repeat: b/sin B = c/sin C. All three ratios are equal.
THE LAW OF SINES, IN ORDER
  • ?In the triangle at A: h = b sin A
  • ?Drop a height h from C to side c, making two right triangles
  • ?In the triangle at B: h = a sin B
  • ?Drop a height from A to bring in c/sin C
  • ?Divide by sin A sin B: a/sin A = b/sin B
  • ?Both equal h, so b sin A = a sin B
WHY THIS EXERCISEThe whole law rests on one shared height. Seeing that is what lets you rebuild it when you forget it.

Derivation three: the Law of Cosines

  1. Drop a height h from B to side b. It lands at a point D, splitting b into two pieces.
  2. In the right triangle at C: the piece CD = a cos C, and h = a sin C.
  3. The other piece AD = b - a cos C.
  4. Pythagoras in the right triangle at A: c² = h² + AD² = (a sin C)² + (b - a cos C)².
  5. Expand: c² = a² sin²C + b² - 2ab cos C + a² cos²C.
  6. Since sin²C + cos²C = 1: c² = a² + b² - 2ab cos C.

The Pythagorean identity from Algebra 2 does the last step. The Law of Cosines is Pythagoras plus a correction for the lean.

Each derivation begins by drawing one extra line from a vertex to the opposite side. That line is a this. Type one word.
The law whose proof ends with sin²C + cos²C = 1 is the Law of this. Type one word.
In the area derivation, h = a sin C replaces h in ½ × b × h. The result is ½ab sin C, with C the this angle. Type one word.

A mixed set

ASA, SAS AND SSS
  • Read the question.
  • Tap your answer.
A triangle ABC with angles A = 75°, B = 40°, C = 65° and sides a = 120, b = ?In the buoy triangle the baseline of 120 m faces the 75° angle at the buoy. Find the side facing the 40° angle, to 1 place.
A triangle ABC with angles A = ?, B = ?, C = 52° and sides a = 8, b = 11, c = ?Ropes of 8 m and 11 m leave the dock corner 52° apart. How far apart are their far ends, to 1 place?
A triangle ABC with angles A = ?, B = ?, C = ? and sides a = 7, b = 9, c = 12Stakes 7, 9 and 12 meters apart. Find the angle across from the 12 m side, to 1 place.
A triangle ABC with angles A = ?, B = ?, C = 52° and sides a = 8, b = 11Two ropes of 8 m and 11 m meet at 52°. What area do they fence with the segment joining their ends, to 1 place?

Which solution makes sense?

For the three stakes, arccos gave 96.4°, an obtuse angle. The Law of Sines could not tell you that: sin 96.4° and sin 83.6° are equal.

So when a triangle might have an obtuse angle, find that angle with the Law of Cosines first. Arccos never hides an obtuse answer.

And if the SSS cosine comes out beyond -1 or 1, the three lengths do not make a triangle at all.

StatementTrue or false?
Area = ½ab sin C uses the angle between sides a and b.?
The Law of Sines proof uses one height written two ways.?
The Law of Cosines proof uses sin²C + cos²C = 1.?
The Law of Sines can tell an obtuse angle from its supplement.?
0.5 × 8 × 11 = 44?
WHY THIS EXERCISEKnowing where each law comes from also tells you where it can mislead you.

Clear reasoning. Tomorrow the laws show up around the Boathouse, and the week comes back in a mixed set.

← Wednesday