The ambiguous sine-rule case and valid triangle counts
Введено| English | Значение |
|---|---|
| ambiguous case/æmˈbɪɡjuːəs keɪs/ | неоднозначный случай |
How do we check whether both triangles exist?
- Two sides and a non-included angle can describe two different plots of land.
- This lesson studies ambiguous case 歧义情形: Given triangle data that can produce two different valid triangles.
The method
- Let A be opposite a, with b another known side.
- The sine rule gives sinB=b sinA/a.
- If this ratio is above 1 there is no triangle.
- Otherwise test both B=arcsin(b sinA/a) and its supplement 180°−B.
- Retain only positive C=180°−A−B; when B=90° the two branches coincide.
- Every side must be positive.
Given A=30°, a=7, b=10, why do TWO valid triangles exist?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The inverse sine supplies an acute candidate, and its supplement is a second candidate. Both survive only because h<a<b with h=b sinA=5; with a=5 exactly one right triangle fits, and with a=4 none does.
A worked case
Take A=30°, a=7 and b=10. Then sinB=5/7, so B≈45.5847° or 134.4153°. Both give positive C, about 104.4153° or 15.5847°. The corresponding third sides are c=5√3+2√6≈13.5592 and c=5√3−2√6≈3.7613. To derive them, place A at the origin and C=(10cos30°,10sin30°)=(5√3,5); B=(c,0). The condition BC=7 gives (c−5√3)²+25=49. Both intersections lie on the positive baseline. Their areas are bc sinA/2=25√3/2±5√6, so the same data does not determine one area.
The ambiguous sine-rule case and valid triangle counts
Let A be opposite a, with b another known side
Justify each triangle candidate or transformed key point against its defining conditions.
How many triangles fit A=30°, a=7 and b=10?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
h=10sin30°=5; 5<7<10 gives two positive baseline intersections.
Where it goes wrong
- A second inverse-sine value is only a candidate: it may make the remaining angle zero or negative.
- For acute A and h=b sinA, the cases are a<h: none; a=h: one right triangle; h<a<b: two; a≥b: one.
- At a=b the extra baseline intersection is c=0, a degenerate shape.
- For right or obtuse A, a must exceed b for one triangle; otherwise none.
Not true: Two inverse-sine values give two valid triangles. The second angle is valid only if it leaves a positive third angle; sometimes only one triangle exists.
How many fit A=30°, a=5 and b=10?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
a=h=5 gives one right triangle; the inverse-sine branches coincide.
The second angle is valid only if it leaves a positive third angle.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. The second angle is valid only if it leaves a positive third angle; sometimes only one triangle exists.
Applying it
- In the example h=5: changing a to 5 gives one right triangle, to 4 gives none, and to 10 gives one with A=B=30°.
- Check the angle sum, side order and cosine-rule reconstruction for every candidate.
- State both configurations when the given information does not select one, and keep exact lengths/areas where possible.
How many fit A=30°, a=4 and b=10?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
a=4<h=5 makes sinB=5/4>1, so no triangle exists.
Match each ambiguous-case condition (acute A, h=b sinA) to its triangle count.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
A supplementary candidate is only a candidate: it must leave a positive third angle and positive sides before it counts.
In your course
- AQA A-level Mathematics (7357), specification section E.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Ambiguous case: Given triangle data that can produce two different valid triangles.
- Let A be opposite a, with b another known side.
- Watch for: A second inverse-sine value is only a candidate: it may make the remaining angle zero or negative.