Question

1. find ac

1. find \\(m\angle a\\)

1. find \\(m\angle b\\)

1. find \\(m\angle c\\)

ii. solve the unknown in each triangle.
1.

4.

2.

5.

Explanation:

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

Here are the step-by-step solutions for the first four problems shown at the top of the worksheet (numbered 3, 4, 5, and 6).

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Problem 3: Find \(AC\)

Step 1: Find the third angle

The sum of angles in a triangle is \(180^\circ\). We find the measure of angle \(C\):

$$ m\angle C = 180^\circ - (83^\circ + 38^\circ) = 180^\circ - 121^\circ = 59^\circ $$

Step 2: Apply the Law of Sines

The Law of Sines relates the lengths of the sides of a triangle to the sines of its angles:

$$ \frac{AC}{\sin(B)} = \frac{AB}{\sin(C)} $$

Substitute the known values:

$$ \frac{AC}{\sin(83^\circ)} = \frac{25}{\sin(59^\circ)} $$

Step 3: Solve for \(AC\)

$$ AC = \frac{25 \cdot \sin(83^\circ)}{\sin(59^\circ)} $$

$$ AC \approx \frac{25 \cdot 0.9925}{0.8572} \approx 28.95\text{ m} $$

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Problem 4: Find \(m\angle A\)

Step 1: Identify the given information

We are given two sides and the included angle (Side-Angle-Side):

• \(a = BC = 7\text{ yd}\)
• \(c = AB = 28\text{ yd}\)
• \(B = 75^\circ\)

We need to find the measure of angle \(A\). First, we must find the length of the opposite side \(b = AC\) using the Law of Cosines.

Step 2: Use the Law of Cosines to find \(b\)

$$ b^2 = a^2 + c^2 - 2ac \cos(B) $$

$$ b^2 = 7^2 + 28^2 - 2(7)(28) \cos(75^\circ) $$

$$ b^2 = 49 + 784 - 392(0.2588) $$

$$ b^2 = 833 - 101.45 = 731.55 $$

$$ b \approx \sqrt{731.55} \approx 27.05\text{ yd} $$

Step 3: Use the Law of Sines to find \(m\angle A\)

$$ \frac{\sin(A)}{a} = \frac{\sin(B)}{b} $$

$$ \frac{\sin(A)}{7} = \frac{\sin(75^\circ)}{27.05} $$

$$ \sin(A) = \frac{7 \cdot \sin(75^\circ)}{27.05} \approx \frac{7 \cdot 0.9659}{27.05} \approx 0.2500 $$

$$ m\angle A = \sin^{-1}(0.2500) \approx 14.48^\circ $$

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Problem 5: Find \(m\angle B\)

Step 1: Identify the given information

We are given:

• Angle \(C = 28^\circ\)
• Side opposite to \(C\): \(c = AB = 32\text{ mi}\)
• Side opposite to \(B\): \(b = AC = 21\text{ mi}\)

Step 2: Apply the Law of Sines

Since we know an angle and its opposite side, we can use the Law of Sines to find angle \(B\):

$$ \frac{\sin(B)}{b} = \frac{\sin(C)}{c} $$

$$ \frac{\sin(B)}{21} = \frac{\sin(28^\circ)}{32} $$

Step 3: Solve for \(m\angle B\)

$$ \sin(B) = \frac{21 \cdot \sin(28^\circ)}{32} $$

$$ \sin(B) \approx \frac{21 \cdot 0.4695}{32} \approx 0.3081 $$

$$ m\angle B = \sin^{-1}(0.3081) \approx 17.94^\circ $$

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Problem 6: Find \(m\angle C\)

Step 1: Identify the given information

We are given:

• Angle \(A = 98^\circ\)
• Side opposite to \(A\): \(a = BC = 19\text{ ft}\)
• Side opposite to \(C\): \(c = AB = 11\text{ ft}\)

Step 2: Apply the Law of Sines

$$ \frac{\sin(C)}{c} = \frac{\sin(A)}{a} $$

$$ \frac{\sin(C)}{11} = \frac{\sin(98^\circ)}{19} $$

Step 3: Solve for \(m\angle C\)

$$ \sin(C) = \frac{11 \cdot \sin(98^\circ)}{19} $$

$$ \sin(C) \approx \frac{11 \cdot 0.9903}{19} \approx 0.5733 $$

$$ m\angle C = \sin^{-1}(0.5733) \approx 34.98^\circ $$

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Answer:

• Problem 3: \(AC \approx 29.0\text{ m}\)
• Problem 4: \(m\angle A \approx 14.5^\circ\)
• Problem 5: \(m\angle B \approx 17.9^\circ\)
• Problem 6: \(m\angle C \approx 35.0^\circ\)