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Chapter 4: Problem 23

The measures of two sides and an angle are given. Determine whether a triangle (or two) exist, and if so, solve the triangle(s). $$\alpha=30^{\circ}, b=18, a=9$$

Short Answer

Expert verified
Only one triangle exists, with \( \beta = 90^{\circ} \) and \( \gamma = 60^{\circ} \).

Step by step solution

01

Determine Applicable Case

Given the values \( a = 9 \), \( b = 18 \), and \( \alpha = 30^{\circ} \), this problem involves two sides and a non-included angle (SSA). Thus, the Law of Sines is applicable, which will help determine the number of possible triangles.
02

Use the Law of Sines

The Law of Sines states that \( \frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} \). Substitute the known values: \( \frac{9}{\sin(30^{\circ})} = \frac{18}{\sin(\beta)} \). Since \( \sin(30^{\circ}) = 0.5 \), the equation becomes \( \frac{9}{0.5} = \frac{18}{\sin(\beta)} \). Simplify this to find \( \sin(\beta) = 1 \), leading to \( \beta = 90^{\circ} \).
03

Check Triangle Existence (Ambiguous Case)

SSA configuration can lead to the ambiguous case. The calculation from Step 2 indicates exactly one triangle because \( \beta = 90^{\circ} \). If \( \beta \) were smaller, a second triangle might exist, but here, only one triangle is possible because angle B is exactly \( 90^{\circ} \).
04

Solve the Triangle

With \( \beta = 90^{\circ} \), we already know two angles. Use the angle sum property of triangles where \( \alpha + \beta + \gamma = 180^{\circ} \). Thus, \( \gamma = 180^{\circ} - 30^{\circ} - 90^{\circ} = 60^{\circ} \). Finally, use the Law of Sines again to find side \( c \): \( \frac{a}{\sin(\alpha)} = \frac{c}{\sin(\gamma)} \) becomes \( \frac{9}{0.5} = \frac{c}{\sin(60^{\circ})} \). Solving this gives \( c = \frac{9 \times \sqrt{3}}{0.5} = 9\sqrt{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Law of Sines
The Law of Sines is a valuable tool in trigonometry, particularly when dealing with triangles. It allows us to find unknown angles or sides of a triangle when given enough information about the other parts.
The formula is expressed as:
  • \( \frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} = \frac{c}{\sin(\gamma)} \)
This means that the ratio of a side of the triangle to the sine of the angle opposite it is the same for all three sides and angles of the triangle.
In the given exercise, we start with two sides and a non-included angle (SSA configuration), which is perfect for using the Law of Sines to solve the triangle. First, solve the known values to find unknown angles.
This can lead us to understand if one triangle or possibly two triangles鈥攁n "ambiguous case"鈥攅xists.
Ambiguous Case
The ambiguous case arises particularly with SSA (Side-Side-Angle) triangle configurations. This situation can sometimes yield more than one valid triangle, which can make solving seem a bit tricky.
This occurs because the given information might satisfy the conditions for more than one triangle configuration, leading to possible multiple solutions.
In our case, solving with the Law of Sines indicates that \( \beta \) equals \( 90^{\circ} \), which uniquely determines the triangle. If the sine value led to another possible angle, such as an obtuse or acute angle plus a matching smaller angle, two triangles could exist.
However, with \( \beta = 90^{\circ} \), the ambiguity is resolved, resulting in a single right triangle.
Triangle Sum Property
Triangle Sum Property is a fundamental aspect of triangles: the sum of all interior angles in any triangle is always \( 180^{\circ} \).
This is a crucial concept when solving triangles since finding one or more angles helps you deduce the rest.
In our exercise: Once \( \beta = 90^{\circ} \) was found using the Law of Sines, the Triangle Sum Property could determine \( \gamma \).
Since \( \alpha + \beta + \gamma = 180^{\circ} \), knowing two angles allows for easy calculation of the third.
Thus, \( \gamma = 180^{\circ} - 30^{\circ} - 90^{\circ} = 60^{\circ} \).
This property is vital in ensuring that our triangle measurements are consistent and accurate, consolidating the rest of the triangle-solving process.

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Most popular questions from this chapter

Calculate \(\csc 40^{\circ}\) in the following two ways: a. Find \(\sin 40^{\circ}\) to three decimal places and then divide 1 by that number. Write this last result to five decimal places. b. With a calculator in degree mode, enter \(40,\) sin, \(1 / \mathrm{x},\) and round the result to five decimal places.

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?

Calculate cot \(54.9^{\circ}\) in the following two ways: a. Find tan \(54.9^{\circ}\) to three decimal places and then divide 1 by that number. Write that number to five decimal places. b. With a calculator in degree mode, enter \(54.9,\) tan, \(1 / \mathrm{x},\) and round the result to five decimal places.

Determine whether each statement is true or false. If you are given the measures of one side and one acute angle of a right triangle, you can solve the right triangle.

Let \(A, B,\) and \(C\) be the lengths of the three sides with \(X, Y,\) and \(Z\) as the corresponding angle measures in a triangle. Write a program using a TI calculator to solve each triangle with the given measures. $$A=\sqrt{12}, B=\sqrt{21}, \text { and } Z=62.8^{\circ}$$
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