Hey there, fellow math adventurers! Ever wonder how those incredible mapping drones navigate treacherous terrain, calculating distances and areas that are impossible to measure directly? Today, we’re diving headfirst into the world of non-right triangles, equipping ourselves with powerful trigonometric tools to solve a high-stakes, real-world challenge!
Imagine this: You’re piloting a state-of-the-art survey drone, high above a jagged alpine region. Your mission? Map a hazardous zone. You’ve pinpointed two critical ground markers, let’s call them Alpha (A) and Bravo (B). From your drone’s current position (Delta, D), you measure the angle formed by these markers, $\angle ADB$, at a crisp $58^\circ$. You also get a precise laser reading to marker A, $AD$, as $1.2$ kilometers.
But here’s the curveball: thick, swirling fog obstructs your direct line of sight to marker B. You can’t measure distance $DB$ directly! Luckily, a quick-thinking ground team at marker A comes through with a crucial measurement: $\angle DAB$ is $40^\circ$.
Your ultimate mission, should you choose to ace it: Determine the elusive distance $DB$ and the total area of the triangular region $ADB$. And, the million-dollar question for any true mathematician: Given this information, is there more than one possible shape this triangle could take?
This isn’t just a theoretical exercise; it’s exactly the kind of problem-solving scenario you’ll encounter in IB Math, and it’s what we’ll conquer by mastering some indispensable trigonometric principles. Ready to fly?
Table of Contents
The Sine Law: Your First Navigator
When you’re dealing with triangles that aren’t nice, neat right-angled ones, the Sine Law is often your first port of call. It’s a fantastic tool for finding unknown sides or angles when you have a specific pattern of information.
Definition: The Sine Law
For any triangle $ABC$ with sides $a, b, c$ opposite angles $A, B, C$ respectively, the Sine Law states:
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$
This means the ratio of a side to the sine of its opposite angle is constant throughout any given triangle.
We typically use the Sine Law when we know:
- Two angles and any one side (often called Angle-Side-Angle (ASA) or Angle-Angle-Side (AAS) cases).
- Two sides and an angle not included between them (Side-Side-Angle (SSA) – warning, this one can be tricky! More on this ‘ambiguous case’ later).
Example: Using the Sine Law
Let’s say we have triangle $RST$, where $\angle R = 75^\circ$, $\angle S = 45^\circ$, and side $s$ (opposite $\angle S$) is $15$ cm. We want to find side $r$ (opposite $\angle R$).
Using the Sine Law, we set up the ratio:
$$ \frac{r}{\sin R} = \frac{s}{\sin S} $$
Substitute the known values:
$$ \frac{r}{\sin 75^\circ} = \frac{15}{\sin 45^\circ} $$
To isolate $r$, multiply both sides by $\sin 75^\circ$:
$$ r = \frac{15 \sin 75^\circ}{\sin 45^\circ} $$
Calculating this out:
$$ r \approx \frac{15 \times 0.9659}{0.7071} \approx 20.48 \, \text{cm} $$
Pretty straightforward, right? As long as you have a full ‘pair’ (an angle and its opposite side), you’re good to go!
Tip: When to Choose Sine Law
The Sine Law is your go-to when you have an ‘angle-side pair’ and you’re trying to find another side or angle where you already know its opposite. If you have two angles, you can always find the third, giving you a full pair!
The Cosine Law: The Pythagorean Power-Up
Sometimes, the Sine Law just isn’t enough. When you don’t have that crucial angle-side pair, or if you’re dealing with specific configurations of sides and angles, the Cosine Law steps in. Think of it as the Pythagorean Theorem’s super-powered cousin, applicable to *all* triangles, not just right-angled ones.
Definition: The Cosine Law
For any triangle $ABC$ with sides $a, b, c$ opposite angles $A, B, C$ respectively, the Cosine Law can be stated in three forms:
- $$ a^2 = b^2 + c^2 – 2bc \cos A $$
- $$ b^2 = a^2 + c^2 – 2ac \cos B $$
- $$ c^2 = a^2 + b^2 – 2ab \cos C $$
The pattern is key: the square of one side equals the sum of the squares of the other two sides, minus two times the product of those two sides, multiplied by the cosine of the angle *opposite* the side you’re solving for.
You’ll pull out the Cosine Law when you know:
- Two sides and the included angle (Side-Angle-Side (SAS)). This is for finding the third side.
- All three sides (Side-Side-Side (SSS)). This is for finding any angle.
Example: Using the Cosine Law (SAS)
Consider triangle $GHI$. Suppose side $g = 9$ cm, side $h = 14$ cm, and the included angle $\angle I = 80^\circ$. We want to find side $i$.
Using the appropriate form of the Cosine Law:
$$ i^2 = g^2 + h^2 – 2gh \cos I $$
Substitute the values:
$$ i^2 = 9^2 + 14^2 – 2(9)(14) \cos 80^\circ $$
$$ i^2 = 81 + 196 – 252 \cos 80^\circ $$
$$ i^2 = 277 – 252(0.1736) $$
$$ i^2 = 277 – 43.72 $$
$$ i^2 = 233.28 $$
Now, take the square root:
$$ i = \sqrt{233.28} \approx 15.27 \, \text{cm} $$
Tip: Rearranging for Angles (SSS)
If you’re given all three sides and need to find an angle, you can rearrange the Cosine Law. For example, to find angle $A$:
$$ \cos A = \frac{b^2 + c^2 – a^2}{2bc} $$
Then, simply use the inverse cosine ($\cos^{-1}$) function to find the angle $A$. This is super handy when you don’t have any angles to start with!
Unlocking Area: From Base & Height to Sine Rule
We all remember the basic area formula for a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. But what happens when you don’t have the perpendicular height? That’s where the Sine Rule for area comes to the rescue! It’s a neat derivation that connects geometry and trigonometry.
Let’s briefly trace its origin. Take any triangle $ABC$. If we drop an altitude (height) $h$ from vertex $C$ to side $AB$ (our base), the standard formula applies. Now, consider the right-angled triangle formed by this altitude. We can use basic trigonometry: $\sin A = \frac{h}{c}$ (where $c$ is the side $AC$). Rearranging this, we get $h = c \sin A$. Substitute this back into our original area formula ($\text{Area} = \frac{1}{2} \times \text{base} \times h$), and voilà!
Rule: Area of a Triangle Using Sine
The area of any triangle is equal to half the product of two sides and the sine of the angle *included* between them. For triangle $ABC$, this gives us three equivalent formulas:
- $$ \text{Area} = \frac{1}{2} bc \sin A $$
- $$ \text{Area} = \frac{1}{2} ac \sin B $$
- $$ \text{Area} = \frac{1}{2} ab \sin C $$
The key here is the ‘Side-Angle-Side’ (SAS) condition: you need two sides and the angle *between* those specific sides.
Example: Calculating Area with the Sine Rule
Imagine triangle $XYZ$, where side $x = 11$ m, side $y = 7$ m, and the included angle $\angle Z = 50^\circ$. Let’s find its area.
Using the formula: $\text{Area} = \frac{1}{2} xy \sin Z $
Substitute the values:
$$ \text{Area} = \frac{1}{2} (11)(7) \sin 50^\circ $$
$$ \text{Area} = 38.5 \sin 50^\circ $$
$$ \text{Area} \approx 38.5 \times 0.7660 $$
$$ \text{Area} \approx 29.50 \, \text{m}^2 $$
Super efficient when you don’t have the height!
The Ambiguous Case: When One Triangle Becomes Two (or Zero!)
Alright, time for a crucial warning sign when using the Sine Law: the Ambiguous Case. This happens specifically when you’re given two sides and a non-included angle (Side-Side-Angle, or SSA). Unlike SAS or SSS, SSA doesn’t always guarantee a unique triangle. Sometimes, you can form two different triangles, or even no triangle at all!
Rule: Conditions for the Ambiguous Case (SSA)
When given Angle $A$, side $a$ (opposite $A$), and side $b$ (adjacent to $A$), the number of possible triangles depends on the relationship between $a, b$, and the altitude $h = b \sin A$.
Assuming $\angle A$ is acute:
- If $a < h$: No triangle can be formed. Side ‘a’ is too short to connect to the base.
- If $a = h$: Exactly one right-angled triangle is formed. Side ‘a’ is precisely the altitude.
- If $h < a < b$: Two distinct triangles can be formed. This is the truly ambiguous part! One will have an acute angle opposite side ‘b’, and the other an obtuse angle (which is supplementary to the acute one).
- If $a \ge b$: Exactly one triangle is formed. Side ‘a’ is long enough to swing out past ‘b’ or meet it directly, preventing a second solution.
Example: Navigating Ambiguity
Let’s consider a triangle where $\angle P = 35^\circ$, side $p = 8$ cm, and side $q = 12$ cm. Is this an ambiguous case?
This is SSA (Angle P, side p opposite it, side q adjacent to it).
First, calculate the altitude $h = q \sin P$:
$$ h = 12 \sin 35^\circ $$
$$ h \approx 12 \times 0.5736 $$
$$ h \approx 6.88 \, \text{cm} $$
Now, compare $p$ (which is $8$ cm) with $h$ ($6.88$ cm) and $q$ ($12$ cm):
We see that $h < p < q$ ($6.88 < 8 < 12$).
This means we are in the ambiguous case! There are two possible triangles that can be formed with these measurements. One will have an acute angle opposite side $q$, and the other will have an obtuse angle (the supplement) opposite side $q$.
Important Note for SSA
Always calculate the altitude ($h = b \sin A$) first when faced with an SSA scenario. This value is your gatekeeper to determining how many solutions exist. If the given angle is obtuse, the rules change slightly, but for IB, focusing on the acute angle conditions is usually sufficient.
IB Exam Challenge: Conquering the Alpine Mapping Mission!
Alright, it’s time to put everything we’ve discussed into action and guide our drone through the Alpine Mapping Challenge!
Recall the mission parameters:
- Drone is at point D.
- Angle $\angle ADB = 58^\circ$.
- Distance $AD = 1.2$ km.
- Ground team measures $\angle DAB = 40^\circ$.
- We need to find distance $DB$ and the area of $\triangle ADB$.
- And, importantly, check for ambiguity.
Solving the Drone Problem
Step 1: Find the third angle, $\angle ABD$.
The sum of angles in a triangle is $180^\circ$.
$\angle ABD = 180^\circ – \angle ADB – \angle DAB$
$\angle ABD = 180^\circ – 58^\circ – 40^\circ = 82^\circ$.
Step 2: Calculate distance $DB$ using the Sine Law.
We now have a known angle-side pair: $AD = 1.2$ km, opposite $\angle ABD = 82^\circ$. We want to find $DB$, which is opposite $\angle DAB = 40^\circ$.
Using the Sine Law:
$$ \frac{DB}{\sin(\angle DAB)} = \frac{AD}{\sin(\angle ABD)} $$
$$ \frac{DB}{\sin 40^\circ} = \frac{1.2}{\sin 82^\circ} $$
$$ DB = \frac{1.2 \sin 40^\circ}{\sin 82^\circ} $$
$$ DB \approx \frac{1.2 \times 0.6428}{0.9903} \approx 0.778 \, \text{km} $$
Step 3: Calculate the Area of $\triangle ADB$.
We now have two sides ($AD = 1.2$ km and $DB \approx 0.778$ km) and the included angle ($\angle ADB = 58^\circ$). This is perfect for the Area Sine Rule!
$$ \text{Area} = \frac{1}{2} \times AD \times DB \times \sin(\angle ADB) $$
$$ \text{Area} = \frac{1}{2} (1.2)(0.778) \sin 58^\circ $$
$$ \text{Area} \approx 0.6 \times 0.778 \times 0.8480 $$
$$ \text{Area} \approx 0.395 \, \text{km}^2 $$
Step 4: Check for the Ambiguous Case.
This is a critical point. The ambiguous case arises only in the SSA (Side-Side-Angle) scenario. Let’s look at what was initially *given* in our problem:
- $\angle ADB = 58^\circ$ (Angle D)
- $AD = 1.2$ km (a side adjacent to Angle D)
- $\angle DAB = 40^\circ$ (Angle A, also adjacent to side AD)
This is an Angle-Side-Angle (ASA) or Angle-Angle-Side (AAS) type of given information. With ASA/AAS, a triangle is always uniquely determined. Therefore, there is no ambiguous case for this problem’s initial setup. There’s only one unique triangle that fits these measurements.
Hypothetical Scenario for Understanding Ambiguity:
If, instead, the problem had been structured as an SSA case, for instance, if you were given:
- $\angle DAB = 40^\circ$ (Angle A)
- $DB = 0.7$ km (Side opposite Angle A)
- $AD = 1.2$ km (Side adjacent to Angle A)
In *this hypothetical* SSA case, we would first calculate the altitude from D to AB:
$$ h = AD \sin(\angle DAB) = 1.2 \sin 40^\circ \approx 1.2 \times 0.6428 \approx 0.771 \, \text{km} $$
Comparing $DB$ ($0.7$ km) with $h$ ($0.771$ km): Since $DB < h$, it would mean that no triangle would exist in this *hypothetical* SSA scenario because side $DB$ is too short to reach the base $AB$. This illustrates why understanding the type of given information is paramount!
Fantastic work! We’ve successfully navigated the high-altitude mapping mission, calculating both the unknown distance and the area, while also developing a keen eye for potential ambiguities.
Conclusion
Today, we’ve powered through the fundamental concepts of non-right triangle trigonometry. You’ve honed your skills with the Sine Law, tackling scenarios with known angle-side pairs. You’ve mastered the Cosine Law, a versatile generalization of the Pythagorean Theorem, for when you have Side-Angle-Side or Side-Side-Side information. We explored the elegant derivation and practical application of the Area formula using the Sine Rule, perfect for situations without a direct height measurement. And crucially, we demystified the Ambiguous Case, learning to identify when multiple triangles are possible (or none at all!) based on the SSA condition and the critical altitude check.
These skills are truly invaluable, not just for crushing your IB Math exams, but for countless real-world applications across surveying, navigation, engineering, and even astronomy. Keep practicing and keep building that confidence!
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.
Additional Practice
Question 1: In $\triangle ABC$, $\angle A = 55^\circ$, $\angle B = 70^\circ$, and side $b = 18$ cm. Find the length of side $a$ to one decimal place.
Show Answer
Solution:
First, find $\angle C = 180^\circ – 55^\circ – 70^\circ = 55^\circ$.
Using the Sine Law: $\frac{a}{\sin A} = \frac{b}{\sin B}$
$\frac{a}{\sin 55^\circ} = \frac{18}{\sin 70^\circ}$
$a = \frac{18 \sin 55^\circ}{\sin 70^\circ} \approx \frac{18 \times 0.819}{0.940} \approx 15.68 \, \text{cm}$
Therefore, $a \approx 15.7$ cm (1 d.p.).
Question 2: A triangular garden plot $DEF$ has sides $d = 12$ m, $e = 15$ m, and the angle between them $\angle F = 48^\circ$. Calculate the length of side $f$ to two decimal places.
Show Answer
Solution:
Using the Cosine Law (SAS): $f^2 = d^2 + e^2 – 2de \cos F$
$f^2 = 12^2 + 15^2 – 2(12)(15) \cos 48^\circ$
$f^2 = 144 + 225 – 360 \cos 48^\circ$
$f^2 = 369 – 360(0.6691)$
$f^2 = 369 – 240.876$
$f^2 = 128.124$
$f = \sqrt{128.124} \approx 11.32 \, \text{m}$
Therefore, $f \approx 11.32$ m (2 d.p.).
Question 3: Calculate the area of a triangle with sides measuring $9$ cm and $13$ cm, and an included angle of $72^\circ$. Give your answer to three significant figures.
Show Answer
Solution:
Using the Area Sine Rule: $\text{Area} = \frac{1}{2} ab \sin C$ (or appropriate variables)
$\text{Area} = \frac{1}{2} (9)(13) \sin 72^\circ$
$\text{Area} = 58.5 \sin 72^\circ$
$\text{Area} \approx 58.5 \times 0.9511$
$\text{Area} \approx 55.649 \, \text{cm}^2$
Therefore, Area $\approx 55.6 \, \text{cm}^2$ (3 s.f.).
Question 4: In $\triangle XYZ$, $x = 10$ km, $y = 17$ km, and $z = 21$ km. Find the measure of $\angle X$ to one decimal place.
Show Answer
Solution:
Using the rearranged Cosine Law for angles:
$\cos X = \frac{y^2 + z^2 – x^2}{2yz}$
$\cos X = \frac{17^2 + 21^2 – 10^2}{2(17)(21)}$
$\cos X = \frac{289 + 441 – 100}{714}$
$\cos X = \frac{630}{714} \approx 0.88235$
$X = \cos^{-1}(0.88235) \approx 28.05^\circ$
Therefore, $\angle X \approx 28.1^\circ$ (1 d.p.).
Question 5: For $\triangle LMN$, $\angle L = 30^\circ$, side $l = 6$ units, and side $m = 10$ units. Determine the number of possible triangles that can be formed and justify your answer.
Show Answer
Solution:
This is an SSA (Side-Side-Angle) case. We need to check for the ambiguous case.
First, calculate the altitude $h = m \sin L$:
$h = 10 \sin 30^\circ = 10 \times 0.5 = 5$ units.
Now compare $l$ (6 units) with $h$ (5 units) and $m$ (10 units):
We have $h < l < m$, which means $5 < 6 < 10$.
According to the conditions for the ambiguous case, if $h < a < b$ (or $h < l < m$ in this case), there are two distinct triangles possible.
Question 6: In $\triangle PQR$, $p = 7$ cm, $q = 11$ cm, and $\angle Q = 60^\circ$. Find the length of side $r$ to one decimal place.
Show Answer
Solution:
This is SSA. We need to find $\angle P$ first using Sine Law:
$\frac{\sin P}{p} = \frac{\sin Q}{q}$
$\frac{\sin P}{7} = \frac{\sin 60^\circ}{11}$
$\sin P = \frac{7 \sin 60^\circ}{11} \approx \frac{7 \times 0.8660}{11} \approx 0.5513$
$P = \sin^{-1}(0.5513) \approx 33.4^\circ$. (Acute solution)
Let’s check for ambiguous case: $h = q \sin P_{given} = 11 \sin 60^\circ \approx 9.53$ (Wait, this isn’t right for SSA check. The $h$ check is for the side *opposite* the given angle. Here $p$ is opposite $P$, and $q$ is opposite $Q$. Given $q$ and $\angle Q$ and $p$. So this is $q$ and $p$ and $\angle Q$. Side $q$ is opposite the angle. $p$ is the other given side. Altitude from P to QR would be $q \sin R$. We are given $Q$, $q$, $p$. So $p < q$ and $p$ is not opposite $Q$. This is Angle $Q$, side $q$, side $p$. So the given angle is $Q$. The side opposite $Q$ is $q$. The other side is $p$. We need to check if $p$ forms an ambiguous case wrt $q$ and $\angle Q$. The altitude $h = p \sin Q = 7 \sin 60^\circ \approx 6.06$. Since $h < q$ ($6.06 < 11$), there is a solution. Also $h < p$ ($6.06 < 7$). Since $h < p$ AND $p < q$ is NOT true ($7 < 11$ IS true), it’s $h < p < q$, which means 2 solutions. Let’s re-evaluate. It’s best to solve for the angle opposite the *other* side first to check ambiguity directly. Find $\angle P$ using Sine Law. $\sin P = 0.5513$. So $\angle P_1 \approx 33.4^\circ$ and $\angle P_2 = 180^\circ – 33.4^\circ = 146.6^\circ$. Both are possible since $p < q$ (7 < 11).
Case 1: $\angle P = 33.4^\circ$. Then $\angle R = 180^\circ – 60^\circ – 33.4^\circ = 86.6^\circ$.
Using Sine Law: $\frac{r}{\sin 86.6^\circ} = \frac{11}{\sin 60^\circ}$
$r = \frac{11 \sin 86.6^\circ}{\sin 60^\circ} \approx \frac{11 \times 0.998}{0.866} \approx 12.67 \, \text{cm}$
Case 2: $\angle P = 146.6^\circ$. Then $\angle R = 180^\circ – 60^\circ – 146.6^\circ = -26.6^\circ$. This is impossible. So only one solution.
My initial ambiguous case check was correct but my reasoning for $h < p < q$ was applied incorrectly to the prompt’s given values. The problem asks to find $r$. If there were two values for $\angle P$, there would be two values for $\angle R$ and thus two values for $r$. Since $\angle P_2$ leads to an invalid angle, only $P_1$ is valid.
Therefore, side $r \approx 12.7$ cm (1 d.p.).
Question 7: An architect is designing a triangular window. The lengths of two sides are $1.5$ m and $2.1$ m, and the angle between them is $110^\circ$. What is the area of the window in square meters, to two decimal places?
Show Answer
Solution:
Using the Area Sine Rule: $\text{Area} = \frac{1}{2} ab \sin C$
$\text{Area} = \frac{1}{2} (1.5)(2.1) \sin 110^\circ$
$\text{Area} = 1.575 \sin 110^\circ$
$\text{Area} \approx 1.575 \times 0.9397$
$\text{Area} \approx 1.480 \, \text{m}^2$
Therefore, Area $\approx 1.48$ m$^2$ (2 d.p.).
Question 8: Three points, X, Y, and Z, are marked on a map. The distance $XY = 8.5$ km, $YZ = 6.2$ km, and $XZ = 10.0$ km. Find the measure of $\angle YXZ$ to the nearest degree.
Show Answer
Solution:
We are given SSS (all three sides), so use the Cosine Law to find an angle. We want $\angle YXZ$, which is $\angle X$. The side opposite $\angle X$ is $YZ$ (or $x$).
$\cos X = \frac{XY^2 + XZ^2 – YZ^2}{2(XY)(XZ)}$
$\cos X = \frac{8.5^2 + 10.0^2 – 6.2^2}{2(8.5)(10.0)}$
$\cos X = \frac{72.25 + 100 – 38.44}{170}$
$\cos X = \frac{133.81}{170} \approx 0.7871$
$X = \cos^{-1}(0.7871) \approx 38.04^\circ$
Therefore, $\angle YXZ \approx 38^\circ$ (nearest degree).
Question 9: In $\triangle RST$, $\angle R = 42^\circ$, side $s = 15$ cm, and side $r = 11$ cm. How many distinct triangles can be formed with these measurements? Justify your answer.
Show Answer
Solution:
This is an SSA case (Angle R, side r opposite it, side s adjacent to it).
Calculate the altitude $h = s \sin R$:
$h = 15 \sin 42^\circ \approx 15 \times 0.6691 \approx 10.0365$ cm.
Now compare $r$ (11 cm) with $h$ (10.0365 cm) and $s$ (15 cm):
We see that $h < r < s$, which means $10.0365 < 11 < 15$.
Because side $r$ is longer than the altitude $h$ but shorter than the other given side $s$, there are two distinct triangles possible.
Question 10: (IB Exam Style Question)
A hiker sets out from camp C and walks 4.5 km on a bearing of $070^\circ$ to point A. From point A, she walks another 6.0 km to point B. The bearing of B from C is $115^\circ$.
- Sketch a diagram showing the points C, A, B, and the given information. Clearly label all known distances and angles.
- Calculate the measure of $\angle BCA$.
- Calculate the distance from C to B (CB).
- Calculate the area of the triangular region CAB.
Reveal Solution
Solution:
Part (a): Sketching the Diagram
(Imagine a sketch with C at the origin, a North line from C. A is 4.5km at 70 degrees from North. Draw a North line from A. B is 6km from A. Draw a North line from C again, B is at 115 degrees from North.)
From the bearings:
- Bearing of A from C is $070^\circ$. So the angle from North at C to CA is $70^\circ$.
- Bearing of B from C is $115^\circ$. So the angle from North at C to CB is $115^\circ$.
- The angle $\angle BCA = 115^\circ – 70^\circ = 45^\circ$.
Part (b): Calculate $\angle BCA$.
As determined from the bearings, $\angle BCA = 115^\circ – 70^\circ = 45^\circ$.
Part (c): Calculate the distance CB.
In $\triangle CAB$, we know side $CA = 4.5$ km, side $AB = 6.0$ km, and $\angle BCA = 45^\circ$. This is an SSA case (Angle C, side AB opposite it, side CA adjacent).
First, find $\angle CBA$ (let’s call it $\angle B$) using the Sine Law:
$\frac{\sin B}{CA} = \frac{\sin C}{AB}$
$\frac{\sin B}{4.5} = \frac{\sin 45^\circ}{6.0}$
$\sin B = \frac{4.5 \sin 45^\circ}{6.0} = \frac{4.5 \times 0.7071}{6.0} \approx 0.5303$
$\angle B_1 = \sin^{-1}(0.5303) \approx 32.0^\circ$
Check for ambiguous case: $\angle B_2 = 180^\circ – 32.0^\circ = 148.0^\circ$.
If $\angle B_2 = 148.0^\circ$, then $\angle CAB = 180^\circ – 45^\circ – 148.0^\circ = -13^\circ$, which is impossible.
So, only one solution exists: $\angle CBA = 32.0^\circ$.
Now, find $\angle CAB = 180^\circ – 45^\circ – 32.0^\circ = 103.0^\circ$.
Finally, find $CB$ using the Sine Law:
$\frac{CB}{\sin(\angle CAB)} = \frac{AB}{\sin(\angle BCA)}$
$\frac{CB}{\sin 103.0^\circ} = \frac{6.0}{\sin 45^\circ}$
$CB = \frac{6.0 \sin 103.0^\circ}{\sin 45^\circ} \approx \frac{6.0 \times 0.9744}{0.7071} \approx 8.26 \, \text{km}$
Distance $CB \approx 8.26$ km.
Part (d): Calculate the area of the triangular region CAB.
We have sides $CA = 4.5$ km, $CB \approx 8.26$ km, and the included angle $\angle BCA = 45^\circ$.
$\text{Area} = \frac{1}{2} (CA)(CB) \sin(\angle BCA)$
$\text{Area} = \frac{1}{2} (4.5)(8.26) \sin 45^\circ$
$\text{Area} \approx 0.5 \times 4.5 \times 8.26 \times 0.7071$
$\text{Area} \approx 13.06 \, \text{km}^2$
Area $\approx 13.1$ km$^2$ (3 s.f.).