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To find the size of the angle \(\theta\) in the triangle with sides of lengths 3, 6, and 8, we can use the Law of Cosines. The Law of Cosines states: \[ c^2 = a^2 + b^2 - 2ab \cos(\theta) \] In this case, \(a = 3\), \(b = 8\), and \(c = 6\). We need to find the angle \(\theta\) opposite the side of length 6. 1. Substitute the given values into the Law of Cosines: \[ 6^2 = 3^2 + 8^2 - 2 \cdot 3 \cdot 8 \cdot \cos(\theta) \] 2. Simplify the equation: \[ 36 = 9 + 64 - 48 \cos(\theta) \] 3. Combine like terms: \[ 36 = 73 - 48 \cos(\theta) \] 4. Isolate \(\cos(\theta)\): \[ 36 - 73 = -48 \cos(\theta) \] \[ -37 = -48 \cos(\theta) \] \[ \cos(\theta) = \frac{37}{48} \] 5. Use the inverse cosine function to find \(\theta\): \[ \theta = \cos^{-1}\left(\frac{37}{48}\right) \] Using a calculator to find the inverse cosine: \[ \theta \approx \cos^{-1}\left(\frac{37}{48}\right) \approx 40^\circ \] Therefore, the size of the angle \(\theta\) to the nearest degree is: \[ \boxed{40^\circ} \]

Question

To find the size of the angle θ\theta in the triangle with sides of lengths 3, 6, and 8, we can use the Law of Cosines. The Law of Cosines states: c2=a2+b2−2abcos⁡(θ) c^2 = a^2 + b^2 - 2ab \cos(\theta) In this case, a=3a = 3, b=8b = 8, and c=6c = 6. We need to find the angle θ\theta opposite the side of length 6. 1. Substitute the given values into the Law of Cosines: 62=32+82−2⋅3⋅8⋅cos⁡(θ) 6^2 = 3^2 + 8^2 - 2 \cdot 3 \cdot 8 \cdot \cos(\theta) 2. Simplify the equation: 36=9+64−48cos⁡(θ) 36 = 9 + 64 - 48 \cos(\theta) 3. Combine like terms: 36=73−48cos⁡(θ) 36 = 73 - 48 \cos(\theta) 4. Isolate cos⁡(θ)\cos(\theta): 36−73=−48cos⁡(θ) 36 - 73 = -48 \cos(\theta) −37=−48cos⁡(θ) -37 = -48 \cos(\theta) cos⁡(θ)=3748 \cos(\theta) = \frac{37}{48} 5. Use the inverse cosine function to find θ\theta: θ=cos⁡−1(3748) \theta = \cos^{-1}\left(\frac{37}{48}\right) Using a calculator to find the inverse cosine: θ≈cos⁡−1(3748)≈40∘ \theta \approx \cos^{-1}\left(\frac{37}{48}\right) \approx 40^\circ Therefore, the size of the angle θ\theta to the nearest degree is: 40∘ \boxed{40^\circ}

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Solution

To find the size of the angle θ\theta in the triangle with sides of lengths 3, 6, and 8, we can use the Law of Cosines. The Law of Cosines states: c2=a2+b2−2abcos⁡(θ) c^2 = a^2 + b^2 - 2ab \cos(\theta) In this case, a=3a = 3, b=8b = 8, and c=6c = 6. We need to find the angle θ\theta opposite the side of length 6.

  1. Substitute the given values into the Law of Cosines: 62=32+82−2⋅3⋅8⋅cos⁡(θ) 6^2 = 3^2 + 8^2 - 2 \cdot 3 \cdot 8 \cdot \cos(\theta)

  2. Simplify the equation: 36=9+64−48cos⁡(θ) 36 = 9 + 64 - 48 \cos(\theta)

  3. Combine like terms: 36=73−48cos⁡(θ) 36 = 73 - 48 \cos(\theta)

  4. Isolate cos⁡(θ)\cos(\theta): 36−73=−48cos⁡(θ) 36 - 73 = -48 \cos(\theta) −37=−48cos⁡(θ) -37 = -48 \cos(\theta) cos⁡(θ)=3748 \cos(\theta) = \frac{37}{48}

  5. Use the inverse cosine function to find θ\theta: θ=cos⁡−1(3748) \theta = \cos^{-1}\left(\frac{37}{48}\right)

Using a calculator to find the inverse cosine: θ≈cos⁡−1(3748)≈40∘ \theta \approx \cos^{-1}\left(\frac{37}{48}\right) \approx 40^\circ

Therefore, the size of the angle θ\theta to the nearest degree is: 40∘ \boxed{40^\circ}

This problem has been solved

Similar Questions

To find the size of the angle \(\theta\) in the right-angled triangle, we can use trigonometric ratios. Given that we have the length of the adjacent side (6) and the hypotenuse (8), we can use the cosine function: \[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6}{8} = \frac{3}{4} \] Now, we need to find the angle \(\theta\) whose cosine is \(\frac{3}{4}\). We can use the inverse cosine function (also known as arccos): \[ \theta = \cos^{-1}\left(\frac{3}{4}\right) \] Using a calculator to find the inverse cosine: \[ \theta \approx \cos^{-1}(0.75) \approx 41.4^\circ\] Therefore, the size of the angle \(\theta\) to one decimal place is: \[ \boxed{41.4^\circ} \]

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