Problem 10 - Trigonometry

Question

For each of the following triangles, solve for \(B\) and use the results to explain why the triangle has the given number of solutions. \(A=20^{\circ}, b=45 \mathrm{~m}, a=25 \mathrm{~m}\); two solutions

Step-by-Step Solution

Answer

Two solutions for \(B\) are possible: \(38.01^{\circ}\) and \(141.99^{\circ}\), resulting in two valid triangles.

Step 1: Identify known angles and sides

We are given the angle \(A = 20^{\circ}\), side \(a = 25\, \text{m}\), and side \(b = 45\, \text{m}\). The task is to find angle \(B\) using the law of sines and determine why there are two solutions.

Step 2: Apply the Law of Sines

Using the law of sines, the formula is given by:\[\frac{a}{\sin A} = \frac{b}{\sin B}\]Substitute the known values to solve for \(\sin B\):\[\frac{25}{\sin 20^{\circ}} = \frac{45}{\sin B}\]Solving for \(\sin B\) gives:\[\sin B = \frac{45 \cdot \sin 20^{\circ}}{25}\]

Step 3: Compute \(\sin B\)

Calculate \(\sin 20^{\circ} \approx 0.3420\), then:\[\sin B = \frac{45 \times 0.3420}{25} = 0.6156\]

Step 4: Determine possible angles \(B\)

Use the inverse sine function to find possible values for \(B\): \[B = \sin^{-1}(0.6156)\]This gives us two potential angles because the sine function is positive in the first and second quadrants:1. \( B_1 = 38.01^{\circ} \)2. \( B_2 = 180^{\circ} - 38.01^{\circ} = 141.99^{\circ} \)

Step 5: Check for valid triangle solutions

Ensure both potential angles \(B\) provide valid triangles. For each angle \(B\), calculate \(C = 180^{\circ} - A - B\): For \(B_1 = 38.01^{\circ}\): \[C_1 = 180^{\circ} - 20^{\circ} - 38.01^{\circ} = 121.99^{\circ}\]For \(B_2 = 141.99^{\circ}\): \[C_2 = 180^{\circ} - 20^{\circ} - 141.99^{\circ} = 18.01^{\circ}\]Both cases result in valid triangles (since all angles are positive and their sum is \(180^{\circ}\)), confirming two solutions.

Key Concepts

Understanding the Law of Sines

The law of sines is an essential concept in trigonometry, particularly useful for solving triangles that are not right-angled. It relates the ratio of the length of a side of a triangle to the sine of its opposite angle. This is especially handy:

• When you know two angles and one side (AAS or ASA scenarios)
• When you know two sides and a non-included angle (SSA scenario, like this exercise)

The law is mathematically represented as:\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\]Here, \(a\), \(b\), and \(c\) are the sides of the triangle, while \(A\), \(B\), and \(C\) are the corresponding opposite angles.

In this exercise, we use the law of sines to solve for angle \(B\) given angle \(A\) and sides \(a\) and \(b\). By rearranging the known values into the formula and solving for \(\sin B\), we obtain a result that is critical in exploring multiple triangle solutions.

Exploring Angle Calculation

Calculating an angle in a triangle can sometimes lead to multiple solutions. This is due to the nature of the sine function, which is positive in both the first and second quadrants. Once you calculate \(\sin B\), you use the inverse sine, \(\sin^{-1}\), to find angle \(B\).

Here, the result \(\sin B = 0.6156\) can correspond to two different angle measures:

• The first solution is the direct angle: \(B_1 = \sin^{-1}(0.6156) \approx 38.01^{\circ}\)
• The second solution uses the property of the sine function, deriving \(B_2 = 180^{\circ} - 38.01^{\circ} \approx 141.99^{\circ}\)

Both calculations are crucial, as they account for the different possible configurations of the triangle when given the same set of side and angle inputs.

Ensuring Valid Triangle Solutions

After discovering possible angles for \(B\), it is important to check if each angle leads to a valid triangle. A triangle is valid if the sum of its angles is exactly \(180^{\circ}\) and all its angles are positive.

For each potential angle \(B\), we compute the third angle \(C\) using:\[C = 180^{\circ} - A - B\]

• For \(B_1 = 38.01^{\circ}\): \(C_1 = 180^{\circ} - 20^{\circ} - 38.01^{\circ} = 121.99^{\circ}\)
• For \(B_2 = 141.99^{\circ}\): \(C_2 = 180^{\circ} - 20^{\circ} - 141.99^{\circ} = 18.01^{\circ}\)

Both sets of angles \((A, B_1, C_1)\) and \((A, B_2, C_2)\) lead to triangles that meet all the criteria of a valid geometric figure. Therefore, this exercise confirms that there are indeed two distinct solutions for the triangle with the given elements.