Question

day 3

d. find the unknown in each triangle. round your answer to the nearest degree.

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

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

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

3.

1. \\(a = 14, c = 20, b = 38^{\circ}\\)

1. \\(a = 60^{\circ}, c = 17, b = 12\\)

Explanation:

🆕 New Concept Discovered: Law of Cosines
Solving non-right triangles using side lengths

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Step 1: Identify the appropriate rule

The given triangles are oblique (non-right) triangles where all three side lengths are known, and we need to find an unknown angle \(\theta\).

When we know all three sides of a triangle (Side-Side-Side, or SSS), we cannot use basic right-triangle trigonometry (SOH CAH TOA). Instead, we use the Law of Cosines:

$$ a^2 = b^2 + c^2 - 2bc \cos(A) $$

To find an angle, we can rearrange this formula to solve for the cosine of the angle:

$$ \cos(\theta) = \frac{\text{side}_1^2 + \text{side}_2^2 - \text{opposite side}^2}{2 \cdot \text{side}_1 \cdot \text{side}_2} $$

Here, the "opposite side" is the side directly across from the angle \(\theta\), while \(\text{side}_1\) and \(\text{side}_2\) are the two sides adjacent to (touching) the angle \(\theta\).

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Step 2: Solve Problem 1

For the first triangle:

• The angle we want to find is \(\theta\).
• The side opposite to \(\theta\) is \(35\text{ mm}\).
• The two adjacent sides are \(39\text{ mm}\) and \(47\text{ mm}\).

Substitute these values into the rearranged Law of Cosines:

$$ \cos(\theta) = \frac{39^2 + 47^2 - 35^2}{2 \cdot 39 \cdot 47} $$

Calculate the terms:

$$ 39^2 = 1521 $$

$$ 47^2 = 2209 $$

$$ 35^2 = 1225 $$

$$ 2 \cdot 39 \cdot 47 = 3666 $$

Substitute back into the equation:

$$ \cos(\theta) = \frac{1521 + 2209 - 1225}{3666} $$

$$ \cos(\theta) = \frac{2505}{3666} \approx 0.6833 $$

Find the inverse cosine to get \(\theta\):

$$ \theta = \arccos(0.6833) \approx 46.89^\circ $$

Rounding to the nearest degree gives:

$$ \theta \approx 47^\circ $$

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Step 3: Solve Problem 2

For the second triangle:

• The angle we want to find is \(\theta\).
• The side opposite to \(\theta\) is \(4.9\text{ m}\).
• The two adjacent sides are \(9.1\text{ m}\) and \(8.3\text{ m}\).

Substitute these values into the formula:

$$ \cos(\theta) = \frac{9.1^2 + 8.3^2 - 4.9^2}{2 \cdot 9.1 \cdot 8.3} $$

Calculate the terms:

$$ 9.1^2 = 82.81 $$

$$ 8.3^2 = 68.89 $$

$$ 4.9^2 = 24.01 $$

$$ 2 \cdot 9.1 \cdot 8.3 = 151.06 $$

Substitute back into the equation:

$$ \cos(\theta) = \frac{82.81 + 68.89 - 24.01}{151.06} $$

$$ \cos(\theta) = \frac{127.69}{151.06} \approx 0.8453 $$

Find the inverse cosine to get \(\theta\):

$$ \theta = \arccos(0.8453) \approx 32.29^\circ $$

Rounding to the nearest degree gives:

$$ \theta \approx 32^\circ $$

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Step 4: Solve Problem 3

For the third triangle:

• The angle we want to find is \(\theta\).
• The side opposite to \(\theta\) is \(13\text{ cm}\).
• The two adjacent sides are \(9.4\text{ cm}\) and \(7\text{ cm}\).

Substitute these values into the formula:

$$ \cos(\theta) = \frac{9.4^2 + 7^2 - 13^2}{2 \cdot 9.4 \cdot 7} $$

Calculate the terms:

$$ 9.4^2 = 88.36 $$

$$ 7^2 = 49 $$

$$ 13^2 = 169 $$

$$ 2 \cdot 9.4 \cdot 7 = 131.6 $$

Substitute back into the equation:

$$ \cos(\theta) = \frac{88.36 + 49 - 169}{131.6} $$

$$ \cos(\theta) = \frac{-31.64}{131.6} \approx -0.2404 $$

(Note: A negative cosine value means the angle is obtuse, or greater than \(90^\circ\).)

Find the inverse cosine to get \(\theta\):

$$ \theta = \arccos(-0.2404) \approx 103.91^\circ $$

Rounding to the nearest degree gives:

$$ \theta \approx 104^\circ $$

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Answer:

1. \(\theta = 47^\circ\)
2. \(\theta = 32^\circ\)
3. \(\theta = 104^\circ\)