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Chapter 5: Problem 32

Use the Law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places. $$A=120^{\circ}, \quad a=25, \quad b=24$$

Short Answer

Expert verified
After calculated, we obtain two potential solutions for the triangle: The first triangle is with angles \(A=120^{\circ}\), \(B1 = 53.13^{\circ}\), \(C1 = 6.87^{\circ}\) and sides \(a=25\), \(b=24\), \(c1=21.79\). The second triangle is with angles \(A=120^{\circ}\), \(B2 = 126.87^{\circ}\), \(C2 = -66.87^{\circ}\) and sides \(a=25\), \(b=24\), \(c2=-71.8\), so it isn't a valid solution. The only solution is the first triangle.

Step by step solution

01

Apply the Law of Sines to find unknown angle.

We can start by using the law of sines to find the unknown angle \(B\). The law of sines states that \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}\), where \(A\), \(a\), \(B\), and \(b\) are the lengths of the sides and measures of the angles of a triangle, respectively. By substituting the given values, we get \(\sin(B)=\frac{b}{a} \sin(A) = \frac{24}{25} \sin(120^{\circ})\).
02

Calculate angle B.

Solving the above equation, we find that \(\sin(B) = 0.8062\). However, since the sine function has two solutions in [0, 180], \(B\) could be in the second quadrant as well. Thus, \(B\) could be either \(B1 = \arcsin(0.8062)\) or \(B2 = 180^{\circ} - \arcsin(0.8062)\).
03

Calculate angle C for each potential triangle.

Knowing that the sum of the angles in any triangle must equal \(180^{\circ}\), we find the third angle for both potential triangles. For \(B1\), \(C1 = 180^{\circ}-A-B1\). For \(B2\), \(C2 = 180^{\circ}-A-B2\). If the calculated \(C\) is positive, then we have a valid triangle solution.
04

Use the Law of Sines to find the third side.

Use the law of sines to find side \(c\) for each potential triangle. We use the same law of sines \(\frac{a}{\sin(A)}=\frac{c}{\sin(C)}\) to find \(c\). After substituting the given values and solving the equation, we find \(c\). For \(C1\), \(c1 = \frac{a}{\sin(A)} \sin(C1)\), and for \(C2\), \(c2 = \frac{a}{\sin(A)} \sin(C2)\).
05

Check if both solutions are valid.

Some of the results can produce a non-existent triangle using the Ambiguous case of the Law of Sines. If the sides of the triangle are less or equal to zero, then the triangle does not exist. Check if all sides \(a\), \(b\), and both \(c1\) and \(c2\) are positive to ensure the triangles exist.

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Key Concepts

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

Triangle Solving
When tackling problems involving a triangle using trigonometric laws, we often need to determine unknown sides or angles.
In this exercise, we aim to solve the triangle using the Law of Sines, given one angle and two sides. The goal is to find all unknown angles and possibly the missing side.
To begin solving a triangle:
  • Identify all given information. Here, an angle and two adjacent sides are provided.
  • Choose a method for finding the missing elements. The Law of Sines is suitable when dealing with angle-side pairs.
  • Understanding the relationship: In a triangle, the sum of the angles will always equal 180掳. This will guide you in calculating any undiscovered angles.
Calculating these elements step by step will provide potential solutions to the triangle, leading you to check for the existence of valid triangles.
Trigonometry
Trigonometry is a branch of mathematics that explores the relationships between the sides and angles of triangles. In this exercise, we applied the Law of Sines, one of trigonometry鈥檚 core tools.
The Law of Sines relates the lengths of the sides of a triangle to the sines of its angles. Effectively, it states:\[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]This relationship is particularly useful when you have information about two angles and one side (AAS or ASA) or two sides and one non-included angle (SSA).
In our case, we used:
  • Known angle: 120掳
  • Known sides: 25 and 24
By substituting the known values into the equation, we solved for the unknown angle and later the missing side using a similar method. The challenge lies in calculating accurately and considering any potential additional solutions due to the sine function's properties.
Ambiguous Case
The Ambiguous Case in trigonometry arises when using the Law of Sines, particularly when given two sides and a non-included angle (SSA) to find unknown parts of a triangle.
In this scenario:
  • The sine function can yield two possible angles for a given sine value. Hence, two different triangles may result.
  • For instance, an angle B could be either very acute or just shy of 180掳, making it obtuse.
You check both potential angle values because:
  • Each leads to different configurations and potentially different side lengths.
  • One or both solutions may be valid based on the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the third side.
Evaluating each outcome carefully ensures that only practical solutions are considered, ruling out impossible triangles.

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Most popular questions from this chapter

Fill in the blanks. When you are given three sides of a triangle, you use the Law of _____ to find the three angles of the triangle.

The height \(h\) (in feet) above ground of a seat on a Ferris wheel at time \(t\) (in minutes) can be modeled by \(h(t)=53+50 \sin \left(\frac{\pi}{16} t-\frac{\pi}{2}\right)\) The wheel makes one revolution every 32 seconds. The ride begins when \(t=0\) (a) During the first 32 seconds of the ride, when will a person on the Ferris wheel be 53 feet above ground? (b) When will a person be at the top of the Ferris wheel for the first time during the ride? If the ride lasts 160 seconds, then how many times will a person be at the top of the ride, and at what times?

The length \(s\) of a shadow cast by a vertical gnomon (a device used to tell time) of height \(h\) when the angle of the sun above the horizon is \(\theta\) can be modeled by the equation \(s=\frac{h \sin \left(90^{\circ}-\theta\right)}{\sin \theta}\) (a) Verify that the expression for \(s\) is equal to \(h \cot \theta\) (b) Use a graphing utility to complete the table. Let \(h=5\) feet. (c) Use your table from part (b) to determine the angles of the sun that result in the maximum and minimum lengths of the shadow. (d) Based on your results from part (c), what time of day do you think it is when the angle of the sun above the horizon is \(90^{\circ} ?\)

A plane flies 810 miles from Franklin to Centerville with a bearing of \(75^{\circ} .\) Then it flies 648 miles from Centerville to Rosemount with a bearing of \(32^{\circ} .\) Draw a figure that visually represents the situation. Then find the straight-line distance and bearing from Franklin to Rosemount.

In Exercise \(64,\) the Law of Cosines was used to solve a triangle in the two- solution case of SSA. Can the Law of cosines be used to solve the no-solution and single-solution cases of SSA? Explain.
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