Problem 68 - Precalculus

Question

Explain the mistake that is made. Solve the triangle \(a=6, b=2,\) and \(c=5\). Solution: Step 1: Find \(\beta\) Apply the Law of Cosines. \(b^{2}=a^{2}+c^{2}-2 a c \cos \beta\) Solve for \(\beta\) \(\beta=\cos ^{-1}\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)\) Let \(a=6, b=2\) \(c=5 . \quad \beta \approx 18^{\circ}\) Step 2: Find \(\alpha\) \(\begin{array}{ll}\text { Apply the Law } & \frac{\sin \alpha}{a}=\frac{\sin \beta}{b} \\ \text { of sines. } & a\end{array}\) Solve for \(\alpha\) \(\alpha=\sin ^{-1}\left(\frac{a \sin \beta}{b}\right)\) Let \(a=6, b=2\) and \(\beta=18^{\circ}\) \(\alpha \approx 68^{\circ}\) Step 3: Find \(\gamma\) \(\alpha+\beta+\gamma=180^{\circ}\) $$ \begin{aligned} 68^{\circ}+18^{\circ}+\gamma &=180^{\circ} \\ \gamma & \approx 94^{\circ} \end{aligned} $$ \(a=6, b=2, c=5, \alpha \approx 68^{\circ}, \beta \approx 18^{\circ},\) and \(\gamma \approx 94^{\circ}\) This is incorrect. The longest side is not opposite the largest angle. What mistake was made?

Step-by-Step Solution

Answer

The mistake is in not correctly ensuring the largest side is opposite the largest angle when calculating, primarily due to incorrect checks after calculation for supplemental consideration.

Step 1: Determining the Incorrect Angle Calculation

To solve the triangle, we need to use the Law of Cosines correctly to determine every angle. Initially, the solution attempted to find angle \( \beta \), which is opposite to side \( b = 2 \). However, the mistake lies in how the Law of Cosines was applied: \[b^2 = a^2 + c^2 - 2ac \cos \beta \]Substituting the given values, we have:\[2^2 = 6^2 + 5^2 - 2 \times 6 \times 5 \times \cos \beta\]Calculating this:\[4 = 36 + 25 - 60 \cos \beta \4 = 61 - 60 \cos \beta \-57 = -60 \cos \beta \\cos \beta = \frac{57}{60} \approx 0.95 \\beta \approx \cos^{-1}(0.95) \approx 18.19^\circ,\]confirming the calculation for \( \beta \) is correct.

Step 2: Identifying the Mistake in the Law of Sines

The solution uses the Law of Sines to find \( \alpha \): \[\frac{\sin \alpha}{a} = \frac{\sin \beta}{b} \]Substituting values gives:\[\frac{\sin \alpha}{6} = \frac{\sin 18.19^\circ}{2} \]This yields:\[\sin \alpha = 6 \times \frac{\sin 18.19^\circ}{2} \\sin \alpha \approx 3 \times \sin 18.19^\circ \approx 0.936\]Finding \( \alpha \):\[\alpha = \sin^{-1}(0.936) \approx 68.14^\circ \]However, the solution should typically consider the ambiguous case where another potential angle could be supplementary, leading to a discrepancy with \( \gamma \). The existing deduction method overestimated \( \alpha \). This requires a checks and balances approach by recalculating given the triangle’s properties.

Step 3: Recalculating the Third Angle

Given the errors in angle calculation process for \( \alpha \) and \( \beta \), the overall triangle check must go back to validating which findings match the logical triangle properties with side lengths. Knowing that the largest angle should be opposite the longest side, our corrected findings point to ensuring \( \gamma \), opposite side \( c=5 \), re-calculate:Using the angles by computation approach:- From \( \alpha \approx 68.14^\circ \) and \( \beta \approx 18.19^\circ \), computing \( \gamma = 180^\circ - 68.14^\circ - 18.19^\circ \approx 93.67^\circ \),- Still shows all values fall around an algebraically equated real triangle, contingent finding a form error for consecutive attempts if miscount still occurs.

Key Concepts

Triangle Solution

Solving a triangle involves finding all the angles and sides when some of them are known. There are various methods to tackle this: the Law of Cosines and the Law of Sines are both extremely useful tools. In the original exercise, we attempted to solve the triangle with sides \(a=6\), \(b=2\), and \(c=5\). The goal was to find the angles \(\alpha\), \(\beta\), and \(\gamma\).
Understanding the relationship between the sides and angles is critical: each side of the triangle is opposite to a specific angle. When solving, we often begin by finding one angle using the Law of Cosines, then proceed to find the remaining angles using supplementary equations like the relation \(\alpha + \beta + \gamma = 180^\circ\).
Always ensure that the largest angle is opposite the longest side. This checks if your calculations make sense for a real, valid triangle.

Angle Calculation

Accurate angle calculation in a triangle is crucial, and any error can lead to wrong conclusions. In our scenario, the exercise initially estimated \(\beta\) using the Law of Cosines. This angle is opposite side \(b\), measuring 2 units. While calculating, make sure to substitute correctly into the Law of Cosines and solve for \(\beta\):

• \(\cos \beta = \frac{a^2 + c^2 - b^2}{2ac}\)

The Law of Cosines provided \(\beta \approx 18^\circ\). To further progress, the Law of Sines was utilized to calculate \(\alpha\):

• \(\frac{\sin \alpha}{a} = \frac{\sin \beta}{b}\)

Make use of the inverse sine function to find the angle from the sine value:

• \(\alpha = \sin^{-1}(\frac{a \sin \beta}{b})\)

This yielded \(\alpha \approx 68^\circ\). Finally, using the angle sum property, determine \(\gamma\) by ensuring it complements \(\alpha + \beta\) to 180 degrees.

Law of Sines

The Law of Sines is a powerful tool in trigonometry, offering a straightforward way to find unknown angles or sides in a triangle. It states that the ratio of a side's length to the sine of its opposite angle is constant across all three sides and angles:

• \(\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}\)

This law is especially helpful when at least one angle and its opposite side are known, enabling the calculation of other angles or sides efficiently. In the exercise, this law helped find \(\alpha\), and ensured the sum \(\alpha + \beta + \gamma = 180^\circ\). Re-checking calculations here with the triangle inequality concept helps confirm that no side or angle is miscalculated, a common oversight when using trigonometric functions.

Triangle Inequality

The triangle inequality theorem offers a fundamental property for triangles: the sum of lengths of any two sides must be greater than the third side. This is important to validate the triangle’s feasibility. In our exercise, we had sides \(a=6\), \(b=2\), and \(c=5\). Confirming:

• \(a + b > c \rightarrow 6 + 2 > 5\)
• \(a + c > b \rightarrow 6 + 5 > 2\)
• \(b + c > a \rightarrow 2 + 5 > 6\)

Each inequality holds true, verifying a valid triangle configuration. Remember that the triangle inequality is not just a check for the sides, but also to cross-confirm against any derived angle measures. If the longest side isn't opposite the largest angle, recalibrate your calculations to fit the criteria or reconsider your trigonometry application.