Problem 24 - Precalculus

Question

Solve each triangle. $$a=11, c=12, \gamma=60^{\circ}$$

Step-by-Step Solution

Answer

\(b \approx 14.11\), \(\alpha \approx 43.09^{\circ}\), \(\beta \approx 76.91^{\circ}\).

Step 1: Use Law of Cosines

To solve the triangle, begin by finding the unknown side \(b\) using the law of cosines: \[b^2 = a^2 + c^2 - 2ac \cos \gamma\]Substitute \(a = 11\), \(c = 12\), and \(\gamma = 60^{\circ}\) into the formula:\[b^2 = 11^2 + 12^2 - 2(11)(12) \cos 60^{\circ}\]\[b^2 = 121 + 144 - 132 \times 0.5\]\[b^2 = 121 + 144 - 66\]\[b^2 = 199\]Now solve for \(b\):\[b = \sqrt{199} \approx 14.11\]

Step 2: Use Law of Sines to Find Angle \(\alpha\)

Next, use the law of sines to find \(\alpha\):\[\frac{a}{\sin \alpha} = \frac{c}{\sin \gamma}\]Substitute \(a = 11\), \(c = 12\), and \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\):\[\frac{11}{\sin \alpha} = \frac{12}{\frac{\sqrt{3}}{2}}\]\[\sin \alpha = \frac{11 \times \frac{\sqrt{3}}{2}}{12} \]\[\sin \alpha = \frac{11\sqrt{3}}{24}\]\[\alpha = \sin^{-1}\left(\frac{11\sqrt{3}}{24}\right) \approx 43.09^{\circ}\]

Step 3: Find Angle \(\beta\)

Finally, find the remaining angle \(\beta\) using the fact that the sum of angles in a triangle is \(180^{\circ}\):\[\beta = 180^{\circ} - \gamma - \alpha \]Substitute \(\gamma = 60^{\circ}\) and \(\alpha \approx 43.09^{\circ}\):\[\beta = 180^{\circ} - 60^{\circ} - 43.09^{\circ} \]\[\beta \approx 76.91^{\circ}\]

Step 4: Conclusion

Now we have all the sides and angles of the triangle:- Side \(b \approx 14.11\)- Angle \(\alpha \approx 43.09^{\circ}\)- Angle \(\beta \approx 76.91^{\circ}\)

Key Concepts

Law of Cosines

The Law of Cosines is a crucial tool in trigonometry for solving triangles, particularly when dealing with non-right triangles. It helps us find unknown sides or angles when other sides and angles are known. The formula is:

• \[b^2 = a^2 + c^2 - 2ac \cos \gamma\]

This is a generalization of the Pythagorean theorem and is very helpful when you know two sides of a triangle and the angle between them. Let's break it down:

• "\(b\)" stands for the side opposite the angle \(\gamma\).
• "\(a\)" and "\(c\)" are the other two sides of the triangle.
• "\(\gamma\)" is the angle between sides \(a\) and \(c\).

The cosine part adjusts for the fact that the triangle isn't a right triangle. Using this, we can solve for side \(b\) in a triangle, like in our original exercise where \(a = 11\), \(c = 12\), and \(\gamma = 60^{\circ}\). By substituting these values into the formula, we obtain:\[b^2 = 11^2 + 12^2 - 2 \times 11 \times 12 \times \cos(60^{\circ})\]Finally, after calculation, solving for \(b\) gives us \(b \approx 14.11\). This formula is an essential strategy in triangle solving.

Law of Sines

The Law of Sines is another powerful formula in trigonometry that helps solve triangles, especially when a combination of angles and their opposite sides are known. This can be written as:

• \[\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}\]

This relationship means that the ratio of a side length to the sine of its opposite angle is constant for all sides and angles in a given triangle. It's especially useful for finding unknown angles of a triangle. Here's how:In our exercise, after finding side \(b\), we use the Law of Sines to find \(\alpha\) because we know \(a = 11\), \(c = 12\), and angle \(\gamma = 60^{\circ}\). Using the formula, we set:\[\frac{11}{\sin \alpha} = \frac{12}{\sin 60^{\circ}}\]Solving for \(\alpha\), we calculate:\[\sin \alpha = \frac{11 \times \frac{\sqrt{3}}{2}}{12}\]By finding the inverse sine, we get \(\alpha \approx 43.09^{\circ}\). The Law of Sines provides a bridge between angles and sides, making it easier to tackle triangle metrics.

Triangle Solving

Triangle solving involves finding all unknown sides and angles in a triangle. This requires a strong understanding of properties like the sum of angles in a triangle equaling \(180^{\circ}\), and familiarity with trigonometric laws like the Law of Cosines and Law of Sines. Once we've determined some parts of a triangle — two sides and the included angle in this case — we can use these laws and properties to find the rest. In the original problem, we calculated side \(b\), angle \(\alpha\), and finally, used the known angles to find the last angle \(\beta\):

• \[\beta = 180^{\circ} - 60^{\circ} - 43.09^{\circ} \approx 76.91^{\circ}\]

This completes the process of triangle solving. With the sides \(a = 11\), \(b \approx 14.11\), and \(c = 12\), and angles \(\alpha \approx 43.09^{\circ}\), \(\beta \approx 76.91^{\circ}\), and \(\gamma = 60^{\circ}\), the triangle is fully solved, demonstrating how effective these methods are when applied in a step-by-step manner.