Hey there, future math legends! Ever found yourself staring at a problem, convinced there’s only one answer, only for math to pull a fast one and reveal two? Today, we’re diving into one of those mind-bending scenarios: the Ambiguous Case of the Sine Law. Imagine a search and rescue drone operator facing a tricky situation. They’re trying to pinpoint a distress beacon’s location from two ground stations. They have some distances and an angle, but here’s the kicker: is there one exact spot for that beacon, or are there multiple possibilities? This isn’t just a fun thought experiment; it’s a classic IB Math challenge, and by the end of this post, you’ll be able to solve it with confidence!
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The Drone Dilemma Decoded: Our IB Problem Hook
Picture this: Our hero, a skilled drone operator, is coordinating a search from two stations, let’s call them Station Alpha (A) and Station Beta (B), precisely 15 kilometers apart. From Station A, their instruments pick up a signal from a distress beacon (let’s label its location P). The angle formed by the line from A to P and the line from A to B is a precise 40 degrees. We also know the beacon is exactly 10 kilometers away from Station B. So, the burning question is: Can our operator pinpoint the beacon’s exact location, or are there two potential spots?
This isn’t just a cool story; it’s the perfect setup for understanding the Ambiguous Case of the Sine Law. It’s a scenario that pops up in surveying, navigation, and, of course, your IB Math exams!
The Sine Law and Its Quirks: What is the Ambiguous Case?
Before we dive into solving our drone problem, let’s refresh our memory on the Sine Rule itself. It’s one of your fundamental tools for tackling non-right-angled triangles:
Definition: The Sine Rule
For any triangle with sides $a, b, c$ and opposite angles $A, B, C$ respectively, the Sine Rule states:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
This rule is incredibly handy when you know a side and its opposite angle, and one other piece of information (another side or another angle).
Now, while the Sine Rule is powerful, it has a ‘quirk’ when you’re given two sides and an angle that is NOT included between them. This is often called the SSA (Side-Side-Angle) case. This is where the Ambiguous Case can sneak in!
Pro Tip: When to Suspect Ambiguity
Always be on high alert for the Ambiguous Case when you’re given an acute angle, the side opposite it, and another side adjacent to the angle (SSA configuration). This is your primary warning sign!
Visualizing the Ambiguous Case
Let’s imagine we’re building a triangle where we know angle $A$, side $b$ (adjacent to $A$), and side $a$ (opposite $A$). We can draw angle $A$ and side $b$ easily. Now, think of side $a$ as a swinging arm from the end of side $b$. How it meets the ‘base’ line determines the number of triangles.
The crucial piece of the puzzle is the altitude (height), $h$, from the vertex opposite the unknown base to the base line. This height is calculated as $h = b \sin A$.
Rule: Four Scenarios for SSA
Given angle $A$ (acute), side $b$ (adjacent), and side $a$ (opposite $A$), consider these possibilities for side $a$ relative to the altitude $h = b \sin A$:
- No Triangle: If $a < h$, side $a$ is too short to reach the base line. No triangle can be formed.
- One Right Triangle: If $a = h$, side $a$ perfectly forms a right angle with the base line. One unique right-angled triangle.
- The Ambiguous Case (Two Triangles): If $h < a < b$, side $a$ is long enough to reach the base line, but short enough that it can swing and intersect at two distinct points. This creates two possible triangles, one with an acute angle at the base and one with its supplementary obtuse angle. This is the Ambiguous Case!
- One Triangle (Non-Ambiguous): If $a \ge b$, side $a$ is long enough to only intersect the base line at one point, or it forms a unique triangle when $a=b$.
The critical conditions for the Ambiguous Case are: 1) The given angle (A) is acute. 2) The side opposite this angle (a) is shorter than the adjacent side (b). 3) The side opposite the given angle (a) is longer than the altitude ($h = b \sin A$).
Solving the Beacon Mystery: A Step-by-Step Approach
Okay, let’s bring it back to our drone operator and the distress beacon. We have a triangle formed by Station A, Station B, and the beacon’s location P. Let’s list what we know:
- Distance AB = 15 km
- Angle A = 40 degrees
- Distance BP = 10 km (this is side ‘a’ in our triangle, opposite Angle A)
Example: Solving the Drone Problem
Our goal is to find angle P. We can set up the Sine Rule as follows:
$$\frac{BP}{\sin A} = \frac{AB}{\sin P}$$
Substitute the known values:
$$\frac{10}{\sin 40^\circ} = \frac{15}{\sin P}$$
Now, let’s isolate $\sin P$:
$$\sin P = \frac{15 \cdot \sin 40^\circ}{10}$$
Calculate the value:
$$\sin P = \frac{15 (0.6428)}{10} \approx \frac{9.642}{10} = 0.9642$$
Next, find the primary angle $P_1$ using the inverse sine function:
$$P_1 = \arcsin(0.9642) \approx 74.65^\circ$$
Checking for Ambiguity:
Before proceeding, let’s confirm if this is indeed an ambiguous case. Our given angle (A) is $40^\circ$, which is acute. The side opposite A (BP) is 10 km. The adjacent side (AB) is 15 km. Since $10 < 15$ (opposite side is shorter than adjacent side), and angle A is acute, this immediately flags a potential ambiguous case!
We should also check the altitude $h = AB \sin A = 15 \sin 40^\circ \approx 15 (0.6428) \approx 9.642$ km. Since $h < BP < AB$ (i.e., $9.642 < 10 < 15$), we definitely have two possible triangles.
Finding the Second Possible Angle for P:
For the ambiguous case, there’s a second angle whose sine is also $0.9642$. This angle is supplementary to $P_1$:
$$P_2 = 180^\circ – P_1 = 180^\circ – 74.65^\circ = 105.35^\circ$$
Solving for Each Triangle:
Now, we need to find the third angle (Angle B) and the remaining side (AP) for both possible triangles.
Triangle 1 (using $P_1 = 74.65^\circ$):
- Angle B1: $B_1 = 180^\circ – A – P_1 = 180^\circ – 40^\circ – 74.65^\circ = 65.35^\circ$
- Side AP1: Using the Sine Rule: $$\frac{AP_1}{\sin B_1} = \frac{BP}{\sin A}$$ $$\frac{AP_1}{\sin 65.35^\circ} = \frac{10}{\sin 40^\circ}$$ $$AP_1 = \frac{10 \cdot \sin 65.35^\circ}{\sin 40^\circ} = \frac{10 (0.9090)}{0.6428} \approx 14.14 \text{ km}$$
Triangle 2 (using $P_2 = 105.35^\circ$):
- Angle B2: $B_2 = 180^\circ – A – P_2 = 180^\circ – 40^\circ – 105.35^\circ = 34.65^\circ$
- Side AP2: Using the Sine Rule: $$\frac{AP_2}{\sin B_2} = \frac{BP}{\sin A}$$ $$\frac{AP_2}{\sin 34.65^\circ} = \frac{10}{\sin 40^\circ}$$ $$AP_2 = \frac{10 \cdot \sin 34.65^\circ}{\sin 40^\circ} = \frac{10 (0.5686)}{0.6428} \approx 8.85 \text{ km}$$
So, our drone operator has indeed found two possible locations for the distress beacon, given the initial information! In the real world, they’d need additional data, perhaps from a third station, or a different type of measurement to pinpoint the exact location. But for your IB exam, knowing how to find both solutions is absolutely key!
Sharpen Your Skills: Additional Practice
You’ve seen the theory and walked through an example. Now it’s your turn to get hands-on and solidify your understanding. Tackle these problems, and remember to always look out for those sneaky ambiguous cases!
Question 1: In $\triangle ABC$, $a = 12$ cm, $b = 15$ cm, and $A = 35^\circ$. Find all possible values for angle $B$.
Show Answer
Solution:
Using the Sine Rule: $$\frac{b}{\sin B} = \frac{a}{\sin A}$$ $$\frac{15}{\sin B} = \frac{12}{\sin 35^\circ}$$ $$\sin B = \frac{15 \cdot \sin 35^\circ}{12} = \frac{15 (0.5736)}{12} \approx 0.7170$$
Since $A = 35^\circ$ (acute) and $a < b$ ($12 < 15$), this is an ambiguous case.
$$B_1 = \arcsin(0.7170) \approx 45.8^\circ$$
$$B_2 = 180^\circ – 45.8^\circ = 134.2^\circ$$
Both angles are valid as $A+B_1 = 35+45.8 = 80.8 < 180$ and $A+B_2 = 35+134.2 = 169.2 < 180$.
So, angle $B$ can be approximately $45.8^\circ$ or $134.2^\circ$.
Question 2: A triangle has sides $x=8$ units, $y=10$ units, and angle $X=50^\circ$. Determine the number of possible triangles and, if applicable, solve for all possible values of angle $Y$.
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Solution:
Given angle $X = 50^\circ$ (acute). Side opposite $X$ is $x=8$. Side adjacent to $X$ is $y=10$. Since $x < y$ ($8 < 10$), this is potentially ambiguous.
Calculate altitude $h = y \sin X = 10 \sin 50^\circ = 10 (0.7660) \approx 7.66$ units.
Since $h < x < y$ ($7.66 < 8 < 10$), there are two possible triangles.
Using the Sine Rule: $$\frac{y}{\sin Y} = \frac{x}{\sin X}$$ $$\frac{10}{\sin Y} = \frac{8}{\sin 50^\circ}$$ $$\sin Y = \frac{10 \cdot \sin 50^\circ}{8} = \frac{10 \cdot 0.7660}{8} \approx 0.9575$$
$$Y_1 = \arcsin(0.9575) \approx 73.2^\circ$$
$$Y_2 = 180^\circ – 73.2^\circ = 106.8^\circ$$
Both are valid: $X+Y_1 = 50+73.2 = 123.2 < 180$; $X+Y_2 = 50+106.8 = 156.8 < 180$.
Possible angles for $Y$ are approximately $73.2^\circ$ and $106.8^\circ$.
Question 3: For $\triangle DEF$, $d=7$ m, $e=6$ m, $D=60^\circ$. How many triangles can be formed?
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Solution:
Given angle $D = 60^\circ$ (acute). Side opposite $D$ is $d=7$. Side adjacent to $D$ is $e=6$.
Since $d \ge e$ ($7 \ge 6$), this is not an ambiguous case. Only one triangle can be formed.
Question 4: In $\triangle PQR$, angle $P = 45^\circ$, side $p = 9$ cm, side $q = 12$ cm. Find the length of side $r$ for all possible triangles.
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Solution:
Given $P = 45^\circ$ (acute), $p = 9$, $q = 12$. Since $p < q$, check for ambiguity.
Altitude $h = q \sin P = 12 \sin 45^\circ = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2} \approx 8.485$ cm.
Since $h < p < q$ ($8.485 < 9 < 12$), there are two possible triangles.
First, find angle $Q$:
$$\frac{q}{\sin Q} = \frac{p}{\sin P}$$ $$\frac{12}{\sin Q} = \frac{9}{\sin 45^\circ}$$ $$\sin Q = \frac{12 \cdot \sin 45^\circ}{9} = \frac{12 (0.7071)}{9} \approx 0.9428$$
$$Q_1 = \arcsin(0.9428) \approx 70.5^\circ$$
$$Q_2 = 180^\circ – 70.5^\circ = 109.5^\circ$$
Triangle 1 (using $Q_1 = 70.5^\circ$):
Angle $R_1 = 180^\circ – P – Q_1 = 180^\circ – 45^\circ – 70.5^\circ = 64.5^\circ$.
Find side $r_1$:
$$\frac{r_1}{\sin R_1} = \frac{p}{\sin P}$$ $$\frac{r_1}{\sin 64.5^\circ} = \frac{9}{\sin 45^\circ}$$ $$r_1 = \frac{9 \cdot \sin 64.5^\circ}{\sin 45^\circ} = \frac{9 (0.9026)}{0.7071} \approx 11.47 \text{ cm}$$
Triangle 2 (using $Q_2 = 109.5^\circ$):
Angle $R_2 = 180^\circ – P – Q_2 = 180^\circ – 45^\circ – 109.5^\circ = 25.5^\circ$.
Find side $r_2$:
$$\frac{r_2}{\sin R_2} = \frac{p}{\sin P}$$ $$\frac{r_2}{\sin 25.5^\circ} = \frac{9}{\sin 45^\circ}$$ $$r_2 = \frac{9 \cdot \sin 25.5^\circ}{\sin 45^\circ} = \frac{9 (0.4305)}{0.7071} \approx 5.48 \text{ cm}$$
The two possible lengths for side $r$ are approximately $11.47$ cm and $5.48$ cm.
Question 5: A triangular piece of land has one side 200m long. From one end of this side, a fence extends at an angle of $55^\circ$. The third corner of the land is known to be 180m from the other end of the 200m side. Find the possible lengths of the fence.
Show Answer
Solution:
Let the 200m side be $AB$. Let the angle at A be $55^\circ$. Let the side from B to the third corner (C) be $BC=180$m. We need to find $AC$.
Given $A = 55^\circ$ (acute). Side opposite A is $BC=180$. Adjacent side is $AB=200$. Since $BC < AB$ ($180 < 200$), it’s potentially ambiguous.
Altitude $h = AB \sin A = 200 \sin 55^\circ = 200 (0.8192) \approx 163.84$ m.
Since $h < BC < AB$ ($163.84 < 180 < 200$), there are two possible triangles.
First, find angle $C$:
$$\frac{BC}{\sin A} = \frac{AB}{\sin C}$$ $$\frac{180}{\sin 55^\circ} = \frac{200}{\sin C}$$ $$\sin C = \frac{200 \cdot \sin 55^\circ}{180} = \frac{200 (0.8192)}{180} \approx 0.9102$$
$$C_1 = \arcsin(0.9102) \approx 65.6^\circ$$
$$C_2 = 180^\circ – 65.6^\circ = 114.4^\circ$$
Triangle 1 (using $C_1 = 65.6^\circ$):
Angle $B_1 = 180^\circ – A – C_1 = 180^\circ – 55^\circ – 65.6^\circ = 59.4^\circ$.
Find side $AC_1$ (length of fence):
$$\frac{AC_1}{\sin B_1} = \frac{BC}{\sin A}$$ $$\frac{AC_1}{\sin 59.4^\circ} = \frac{180}{\sin 55^\circ}$$ $$AC_1 = \frac{180 \cdot \sin 59.4^\circ}{\sin 55^\circ} = \frac{180 (0.8607)}{0.8192} \approx 189.0 \text{ m}$$
Triangle 2 (using $C_2 = 114.4^\circ$):
Angle $B_2 = 180^\circ – A – C_2 = 180^\circ – 55^\circ – 114.4^\circ = 10.6^\circ$.
Find side $AC_2$ (length of fence):
$$\frac{AC_2}{\sin B_2} = \frac{BC}{\sin A}$$ $$\frac{AC_2}{\sin 10.6^\circ} = \frac{180}{\sin 55^\circ}$$ $$AC_2 = \frac{180 \cdot \sin 10.6^\circ}{\sin 55^\circ} = \frac{180 (0.184) \text{ (approx)}}{0.8192} \approx 40.4 \text{ m}$$
The possible lengths of the fence are approximately $189.0$m and $40.4$m.
Question 6: In $\triangle XYZ$, $X=30^\circ$, $y=20$ km, $x=9$ km. Calculate the number of possible triangles and explain why.
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Solution:
Given angle $X = 30^\circ$ (acute). Side opposite $X$ is $x=9$. Side adjacent to $X$ is $y=20$.
Since $x < y$ ($9 < 20$), we need to check the altitude.
Altitude $h = y \sin X = 20 \sin 30^\circ = 20 (0.5) = 10$ km.
Since $x < h$ ($9 < 10$), side $x$ is too short to reach the base line. Therefore, no triangle can be formed.
Question 7: An observer at point A measures the angle to a distant landmark L as $42^\circ$. Another observer at point B, 500m away from A, measures the distance to L as 350m. How many possible locations are there for landmark L? Justify your answer.
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Solution:
Let’s form $\triangle ABL$. We are given $A=42^\circ$ (acute), $AB=500$m (adjacent side), and $BL=350$m (opposite side to A).
Since $BL < AB$ ($350 < 500$), this is potentially an ambiguous case.
Calculate altitude $h = AB \sin A = 500 \sin 42^\circ = 500 (0.6691) \approx 334.55$ m.
Compare $BL$ with $h$: $350 > 334.55$. Also $BL < AB$ ($350 < 500$).
Since $h < BL < AB$, there are two possible locations for landmark L.
Question 8: Triangle $ABC$ has $AB=7$, $BC=4$, and angle $A=62^\circ$. Determine angle $C$. If there are multiple solutions, find all of them. If none, state why.
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Solution:
Given $A=62^\circ$ (acute). Side opposite $A$ is $BC=4$. Side adjacent is $AB=7$.
Since $BC < AB$ ($4 < 7$), we check the altitude.
Altitude $h = AB \sin A = 7 \sin 62^\circ = 7 (0.8829) \approx 6.18$ units.
Since $BC < h$ ($4 < 6.18$), side $BC$ is too short to form a triangle. Therefore, no solution for angle $C$ exists; no such triangle can be formed.
Question 9: A ship leaves port P and sails 150 km to point Q. From Q, it sails towards a lighthouse L. The angle PQL is $110^\circ$. The lighthouse L is 200 km from port P. Find the distance QL.
Show Answer
Solution:
We have $\triangle PQL$. We know $PQ=150$ km, $PL=200$ km, and angle $PQL = 110^\circ$.
Here, the given angle ($110^\circ$) is obtuse. The ambiguous case only applies when the given angle is acute. Therefore, there is only one unique triangle possible.
Using the Sine Rule to find angle $L$ (opposite side $PQ$):
$$\frac{PQ}{\sin L} = \frac{PL}{\sin PQL}$$ $$\frac{150}{\sin L} = \frac{200}{\sin 110^\circ}$$ $$\sin L = \frac{150 \cdot \sin 110^\circ}{200} = \frac{150 (0.9397)}{200} \approx 0.7048$$
$$L = \arcsin(0.7048) \approx 44.8^\circ$$
Now find angle $P$:
$$P = 180^\circ – PQL – L = 180^\circ – 110^\circ – 44.8^\circ = 25.2^\circ$$
Finally, find distance $QL$ (side opposite $P$):
$$\frac{QL}{\sin P} = \frac{PL}{\sin PQL}$$ $$\frac{QL}{\sin 25.2^\circ} = \frac{200}{\sin 110^\circ}$$ $$QL = \frac{200 \cdot \sin 25.2^\circ}{\sin 110^\circ} = \frac{200 (0.4258)}{0.9397} \approx 90.6 \text{ km}$$
The distance $QL$ is approximately $90.6$ km.
Question 10 (IB Exam Style Question – Paper 2): A surveyor is trying to locate a hidden marker, M, from two known points, P and Q. The distance PQ is 10 km. From point P, the angle to M (angle QPM) is $30^\circ$. The distance from point Q to the marker M (QM) is 6 km.
- Draw a clear diagram showing the given information.
- Show that there are two possible locations for the marker M.
- Calculate the two possible distances from P to M (PM).
- If additional geological data indicates that angle PMQ is obtuse, determine the distance from P to M in this specific scenario.
Reveal Solution
Solution (IB Exam Style Question):
Diagram: (Imagine a diagram with P at one end, Q some distance away, and two possible positions for M swinging from Q to intersect the ray from P at $30^\circ$. Angle P = $30^\circ$, PQ = 10km, QM = 6km.)
(In an actual exam, you would draw this. Label vertices P, Q, M. Angle QPM is $30^\circ$. Side PQ is $m=10$km. Side QM is $p=6$km.)
Show two possible locations for the marker M:
We are given Angle $P = 30^\circ$ (acute). The side opposite angle $P$ is $QM = 6$ km. The side adjacent to angle $P$ is $PQ = 10$ km.
This is an SSA (Side-Side-Angle) case where the given angle is acute and the side opposite it ($p=6$) is shorter than the adjacent side ($m=10$). This strongly suggests an ambiguous case.
Let’s calculate the altitude $h$ from Q to the line containing PM:
$$h = PQ \sin P = 10 \sin 30^\circ = 10 (0.5) = 5 \text{ km}$$
Now, compare $QM$ with $h$ and $PQ$:
$$h < QM < PQ$$
$$5 < 6 < 10$$
Since $h < QM < PQ$, there are indeed two possible locations for the marker M.
Calculate the two possible distances from P to M (PM):
First, use the Sine Rule to find angle $M$ (angle PMQ, opposite side $PQ$):
$$\frac{QM}{\sin P} = \frac{PQ}{\sin M}$$ $$\frac{6}{\sin 30^\circ} = \frac{10}{\sin M}$$ $$\sin M = \frac{10 \cdot \sin 30^\circ}{6} = \frac{10 (0.5)}{6} = \frac{5}{6} \approx 0.8333$$
Now, find the two possible values for angle $M$:
$$M_1 = \arcsin(\frac{5}{6}) \approx 56.44^\circ$$
$$M_2 = 180^\circ – 56.44^\circ = 123.56^\circ$$
Next, find the corresponding angle $Q$ (angle PQM) for each case:
- Case 1 (Acute M): $Q_1 = 180^\circ – P – M_1 = 180^\circ – 30^\circ – 56.44^\circ = 93.56^\circ$
- Case 2 (Obtuse M): $Q_2 = 180^\circ – P – M_2 = 180^\circ – 30^\circ – 123.56^\circ = 26.44^\circ$
Finally, use the Sine Rule to find $PM$ (which is side $q$):
$$\frac{PM}{\sin Q} = \frac{QM}{\sin P}$$
- For Triangle 1 ($Q_1 = 93.56^\circ$): $$\frac{PM_1}{\sin 93.56^\circ} = \frac{6}{\sin 30^\circ}$$ $$PM_1 = \frac{6 \cdot \sin 93.56^\circ}{\sin 30^\circ} = \frac{6 (0.9981)}{0.5} \approx 11.977 \text{ km}$$
- For Triangle 2 ($Q_2 = 26.44^\circ$): $$\frac{PM_2}{\sin 26.44^\circ} = \frac{6}{\sin 30^\circ}$$ $$PM_2 = \frac{6 (0.4453)}{0.5} \approx 5.344 \text{ km}$$
The two possible values for the distance from P to M are approximately $11.98$ km and $5.34$ km.
If additional geological data indicates that angle PMQ is obtuse, determine the distance from P to M in this specific scenario.
Angle PMQ refers to angle $M$. From part (c), we found two possible angles for $M$: $M_1 \approx 56.44^\circ$ (acute) and $M_2 \approx 123.56^\circ$ (obtuse).
Therefore, if angle PMQ is obtuse, we must use the second case, where $M_2 = 123.56^\circ$. The corresponding distance from P to M for this case is $PM_2 \approx 5.344$ km.
The distance PM in this specific scenario is approximately $5.34$ km.
You’ve now mastered the Ambiguous Case of Sine Law! Remember, it’s that tricky situation in the Side-Side-Angle scenario when the given angle is acute, and the side opposite it is shorter than the adjacent side, but long enough to form two different triangles.
Always be on the lookout for these conditions, as they’ll tell you to find two possible solutions. Just like our drone operator, real-world problems often have multiple possibilities until more information clarifies the situation.
Keep practicing these problems to build your confidence and analytical skills.
Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.