Question

1. \\(\theta = \underline{\quad}\\)

e. solve the triangles. round the measure of sides to nearest tenth and the measure of the angle to the nearest degree.

1.

2.

Explanation:

🆕 New Concept Discovered: Laws of Sines and Cosines
Solving non-right triangles using side and angle relationships.

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Understanding the Tools

When solving non-right (oblique) triangles, we cannot use basic right-triangle trigonometry (\(\text{SOH CAH TOA}\)) directly. Instead, we use two powerful tools:

1. **The Law of Sines**: Useful when we know an angle and its opposite side, plus one other piece of information.
\[ \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c} \]
2. **The Law of Cosines**: Useful when we know two sides and the angle between them (SAS), or all three sides (SSS).
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \quad \text{or} \quad \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} \]

Let's solve each problem step-by-step.

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Part 1: Top Problem (labeled "2.")

We are given a triangle with three known sides:
* Side opposite to \(\theta\): \(a = 4.9\text{ m}\)
* Adjacent sides: \(b = 8.3\text{ m}\) and \(c = 9.1\text{ m}\)

We need to find the angle \(\theta\).

Step 1: Apply the Law of Cosines

Since we know all three sides (SSS), we use the Law of Cosines to find the angle \(\theta\) opposite to the side of length \(4.9\text{ m}\):
\[ 4.9^2 = 8.3^2 + 9.1^2 - 2(8.3)(9.1)\cos(\theta) \]

Step 2: Calculate the values

Calculate the squares and products:
\[ 24.01 = 68.89 + 82.81 - 151.06\cos(\theta) \]
\[ 24.01 = 151.7 - 151.06\cos(\theta) \]

Isolate \(\cos(\theta)\):
\[ -127.69 = -151.06\cos(\theta) \]
\[ \cos(\theta) = \frac{-127.69}{-151.06} \approx 0.8453 \]

Step 3: Find the angle \(\theta\)

Take the inverse cosine:
\[ \theta = \cos^{-1}(0.8453) \approx 32.3^{\circ} \]

Rounding to the nearest degree:
\[ \theta \approx 32^{\circ} \]

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Part E, Problem 1: Triangle ABC (left)

We are given:
* Side \(c = 8\) (opposite to angle \(C\))
* Side \(a = 18\) (opposite to angle \(A\))
* Angle \(C = 25^{\circ}\)

We need to "solve the triangle," which means finding all remaining sides and angles: angle \(A\), angle \(B\), and side \(b\).

Step 1: Find Angle \(A\) using the Law of Sines

Since we know side \(c\) and its opposite angle \(C\), we can set up the ratio:
\[ \frac{\sin(A)}{a} = \frac{\sin(C)}{c} \]
\[ \frac{\sin(A)}{18} = \frac{\sin(25^{\circ})}{8} \]
\[ \sin(A) = \frac{18 \cdot \sin(25^{\circ})}{8} \approx \frac{18 \cdot 0.4226}{8} \approx 0.9509 \]

Since side \(a\) (\(18\)) is longer than side \(c\) (\(8\)), angle \(A\) must be larger than angle \(C\) (\(25^{\circ}\)). Looking at the diagram, angle \(A\) is clearly an obtuse angle (greater than \(90^{\circ}\)).

Let's find both possible angles for \(\sin(A) = 0.9509\):
* Acute option: \(A \approx \sin^{-1}(0.9509) \approx 72^{\circ}\)
* Obtuse option: \(A \approx 180^{\circ} - 72^{\circ} = 108^{\circ}\)

Based on the visual representation of the obtuse angle \(A\):
\[ A \approx 108^{\circ} \]

Step 2: Find Angle \(B\)

The sum of angles in a triangle is always \(180^{\circ}\):
\[ B = 180^{\circ} - A - C \]
\[ B = 180^{\circ} - 108^{\circ} - 25^{\circ} = 47^{\circ} \]

Step 3: Find Side \(b\) using the Law of Sines

\[ \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]
\[ \frac{b}{\sin(47^{\circ})} = \frac{8}{\sin(25^{\circ})} \]
\[ b = \frac{8 \cdot \sin(47^{\circ})}{\sin(25^{\circ})} \approx \frac{8 \cdot 0.7314}{0.4226} \approx 13.8 \]

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Part E, Problem 1 (right): Triangle ABC

We are given:
* Side \(b = 4\) (opposite to angle \(B\))
* Side \(a = 9\) (opposite to angle \(A\))
* Angle \(C = 128^{\circ}\) (included angle between sides \(a\) and \(b\))

We need to find side \(c\), angle \(A\), and angle \(B\).

Step 1: Find Side \(c\) using the Law of Cosines

Since we have two sides and the included angle (SAS):
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
\[ c^2 = 9^2 + 4^2 - 2(9)(4)\cos(128^{\circ}) \]
\[ c^2 = 81 + 16 - 72(-0.6157) \]
\[ c^2 = 97 + 44.33 = 141.33 \]
\[ c = \sqrt{141.33} \approx 11.9 \]

Step 2: Find Angle \(B\) using the Law of Sines

\[ \frac{\sin(B)}{b} = \frac{\sin(C)}{c} \]
\[ \frac{\sin(B)}{4} = \frac{\sin(128^{\circ})}{11.9} \]
\[ \sin(B) = \frac{4 \cdot \sin(128^{\circ})}{11.9} \approx \frac{4 \cdot 0.7880}{11.9} \approx 0.2649 \]
\[ B = \sin^{-1}(0.2649) \approx 15^{\circ} \]

Step 3: Find Angle \(A\)

\[ A = 180^{\circ} - C - B \]
\[ A = 180^{\circ} - 128^{\circ} - 15^{\circ} = 37^{\circ} \]

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Part E, Problem 2: Triangle ABC (bottom-left)

We are given three sides (SSS):
* Side \(a = 22\) (opposite to angle \(A\))
* Side \(b = 20\) (opposite to angle \(B\))
* Side \(c = 16\) (opposite to angle \(C\))

We need to find all three angles: \(A\), \(B\), and \(C\).

Step 1: Find Angle \(A\) using the Law of Cosines

\[ a^2 = b^2 + c^2 - 2bc \cos(A) \]
\[ 22^2 = 20^2 + 16^2 - 2(20)(16)\cos(A) \]
\[ 484 = 400 + 256 - 640\cos(A) \]
\[ 484 = 656 - 640\cos(A) \]
\[ -172 = -640\cos(A) \]
\[ \cos(A) = \frac{-172}{-640} \approx 0.2688 \]
\[ A = \cos^{-1}(0.2688) \approx 74^{\circ} \]

Step 2: Find Angle \(B\) using the Law of Sines

\[ \frac{\sin(B)}{b} = \frac{\sin(A)}{a} \]
\[ \frac{\sin(B)}{20} = \frac{\sin(74.4^{\circ})}{22} \]
\[ \sin(B) = \frac{20 \cdot \sin(74.4^{\circ})}{22} \approx \frac{20 \cdot 0.9632}{22} \approx 0.8756 \]
\[ B = \sin^{-1}(0.8756) \approx 61^{\circ} \]

Step 3: Find Angle \(C\)

\[ C = 180^{\circ} - A - B \]
\[ C = 180^{\circ} - 74^{\circ} - 61^{\circ} = 45^{\circ} \]

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Answer:

Top Problem (2):

• \(\theta = 32^{\circ}\)

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Section E Solutions:

1. Left Triangle:

• \(A = 108^{\circ}\)
• \(B = 47^{\circ}\)
• \(b = 13.8\)

1. Right Triangle:

• \(c = 11.9\)
• \(A = 37^{\circ}\)
• \(B = 15^{\circ}\)

2. Bottom-Left Triangle:

• \(A = 74^{\circ}\)
• \(B = 61^{\circ}\)
• \(C = 45^{\circ}\)