SME
Grade 10 Math · Problem type
The ambiguous case (SSA)
Use this for “A = 30°, a = 6 and b = 10. How many triangles fit, and what is B?”
Diagram
Two sides and an angle not between them.
Two triangles fit: B = 56.44° or 123.56°, c = 11.98 or 5.34 (dashed).
Law of sines: a/sin A = b/sin B = c/sin C ≈ 12
This is an example. Type your own values to start a new problem.
a
sin(
°)
=
b
sin(
°)
h
Height b × sin A
worked out (not typed)
n
Triangles that fit
worked out (not typed)
B₂
Second B, 180° − B
worked out (not typed) · °
C
Angle C
worked out (not typed) · °
C₂
Second C
worked out (not typed) · °
c
Side c
worked out (not typed)
c₂
Second c
worked out (not typed)
Formulas
h
=
b
×
sin(
A
°)
5 = 10 × sin(30°)
n
=
the
triangles
side
a
makes
with
height
h
and
side
b
2 = the triangles side 6 makes with height 5 and side 10
B
=
sin⁻¹(
b
×
sin(
A
°)
a
)
56.4427
=
sin⁻¹(10
×
sin(30°)
6
)
B₂
=
180
−
B
123.5573 = 180 − 56.4427
C
=
180
−
A
−
B
93.5573 = 180 − 30 − 56.4427
C₂
=
180
−
A
−
B₂
26.4427 = 180 − 30 − 123.5573
c
=
a
×
sin(
C
°)
sin(
A
°)
11.9769
=
6
×
sin(93.5573°)
sin(30°)
c₂
=
a
×
sin(
C₂
°)
sin(
A
°)
5.3436
=
6
×
sin(26.4427°)
sin(30°)
Assumptions
•
Side a is across from the acute angle A, and b is the other side next to A.
•
Acute A: a < b × sin A fits none, a = b × sin A or a ≥ b fits one, anything between fits two.
•
Two triangles share the same sin B: B and 180° − B.
Step-by-step
WE KNOW
a = 6, b = 10, A = 30°
FIND
height b × sin A (h), triangles that fit (n), angle B (B), second B, 180° − B (B₂), angle C (C), second C (C₂), side c (c), second c (c₂)
Step 1 · Find height b × sin A (h)
Use h = b × sin(A°)
The
height
from
C
to
the
line
of
side
c
:
b
is
its
hypotenuse.
h
=
10
×
sin(30°)
h
=
10
×
0.5
h
=
5
Step 2 · Find triangles that fit (n)
Use n = the triangles side a makes with height h and side b
a
is
between
h
and
b
:
it
reaches
the
base
line
in
two
places,
so
two
fit.
n
=
2
(
h
<
a
<
b
:
5
<
6
<
10)
Step 3 · Find angle B (B)
Use B = sin⁻¹(b × sin(A°) ÷ a)
Law
of
sines,
solved
for
sin
B
,
then
the
inverse
sine.
B
=
sin⁻¹(10
×
sin(30°)
6
)
B
=
56.4427°
Step 4 · Find second B, 180° − B (B₂)
Use B₂ = 180 − B
The same sine fits the obtuse angle too.
B₂
=
180
−
56.4427
B₂
=
123.5573°
Step 5 · Find angle C (C)
Use C = 180 − A − B
The angles of a triangle add to 180°.
C
=
180
−
30
−
56.4427
C
=
93.5573°
Step 6 · Find second C (C₂)
Use C₂ = 180 − A − B₂
The angles of the second triangle add to 180° too.
C₂
=
180
−
30
−
123.5573
C₂
=
26.4427°
Step 7 · Find side c (c)
Use c = a × sin(C°) ÷ sin(A°)
Law
of
sines:
c
sin
C
=
a
sin
A
.
c
=
6
×
sin(93.5573°)
sin(30°)
c
=
6
×
0.9981
0.5
c
=
5.9886
0.5
c
=
11.9769
Step 8 · Find second c (c₂)
Use c₂ = a × sin(C₂°) ÷ sin(A°)
Law
of
sines
again,
with
the
second
triangle’s
angle
C₂
.
c₂
=
6
×
sin(26.4427°)
sin(30°)
c₂
=
6
×
0.4453
0.5
c₂
=
2.6718
0.5
c₂
=
5.3436
Check
Put the numbers back in. Both sides should match.
5 = 10 × sin(30°) ✓
2 = 2 ✓
56.4427
=
sin⁻¹(10
×
sin(30°)
6
)
✓
123.5573 = 180 − 56.4427 ✓
93.5573 = 180 − 30 − 56.4427 ✓
26.4427 = 180 − 30 − 123.5573 ✓
11.9769
≈
6
×
sin(93.5573°)
sin(30°)
✓
5.3436
≈
6
×
sin(26.4427°)
sin(30°)
✓
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