Friday
Friday: Triangles Around the Boathouse
Friday the Boathouse wall holds a row of triangles, each with its numbers.
"Buoy from stake A, 112.6 meters," Comet reads. "Buoy to post, 8.8. Stake angle, 96.4 degrees."
"Three different setups," Wren says. "What do you notice they share?"
"None had a right angle," Comet says. "And none needed one. We made our own by dropping a height."
Nova sorts the triangles into two rows. "Would you like a hint? One row started with a pair, the other did not."
"Side with its own angle: sines. Otherwise: cosines," Wren says. "Today we read every triangle that way."
"And then the one I care about," Comet says. "How far out is the buoy from the beach, straight back?"
Where the crew used the laws this week
| Question | Case | Law | Answer |
|---|---|---|---|
| distance from stake A to the buoy | ASA | Sines | 112.6 m |
| distance from stake B to the buoy | ASA | Sines | 79.9 m |
| buoy to post, from two ropes | SAS | Cosines | 8.8 m |
| angle at the three stakes | SSS | Cosines | 96.4° |
| chair to the lawn baseline | height | side × sine | 10.3 m |
The buoy's distance from the baseline
The buoy is 112.6 m from stake A, and angle A is 40°. Drop the height from the buoy to the tape.
That height is 112.6 × sin 40° = 112.6 × 0.643 ≈ 72.4 meters. The buoy floats about 72.4 m straight out from the beach.
Check from the other end: 79.9 × sin 65° = 79.9 × 0.906 ≈ 72.4. The two heights agree within rounding.
The triangle's area is ½ × 120 × 112.6 × sin 40° ≈ 4342.7 square meters. That matches ½ × base × height.
Everyday triangles without a right angle
- Two people on a shore sight the same boat from the ends of a tape. The Law of Sines finds its distance.
- A hinged ramp meets a wall at a known angle; its two lengths and the gap fit the Law of Cosines.
- Three posts in a yard: the three distances give every angle through SSS.
- A kite's two strings from two holders and the ground gap between them make a triangle to solve.
Which answer makes sense?
Solving the buoy triangle gives 112.6 m and 79.9 m. The biggest side must face the biggest angle, C = 75°, and 120 does.
If a computed side came out longer than the baseline while facing a smaller angle, a label slipped. Check the pairing.
- Read the question.
- Tap your answer.
The buoy triangle has sides 120 m and 112.6 m with a 40° angle between them. What is its area, to 1 place?
A right triangle has legs 6 m and 8 m. Use the Law of Cosines with C = 90° to find the third side, to 1 place.| Week review | True or false? |
|---|---|
| a/sin A = b/sin B = c/sin C in every triangle. | ? |
| The Law of Cosines needs a right angle. | ? |
| Area = ½ab sin C comes from dropping a height and writing h = a sin C. | ? |
| SSS is solved by the Law of Sines first. | ? |
| The distance from a vertex straight to the opposite side is side × sine of the angle at the end. | ? |
A full week of triangles solved from the shore. Tomorrow is Dock Day, and the family measures across something they cannot cross.