Question

design a problem that could be solved using trigonometry and some of the information from your triangle. for example, given two angles and a side, find the length of the other two sides. once you’ve finished designing your problem, solve it. this is a self - assessment, which will help you to: - assess and evaluate your own work - determine where you are in your learning, where you need to go, and how best to get there - submit a copy of your solutions and self - marked rubric score since you are the assessor, you will provide the feedback using the following rubric.

Explanation:

Designing the Trigonometry Problem

Let's consider a right - triangle problem. Suppose we have a right - triangle \(ABC\) with a right angle at \(C\). We know that angle \(A = 30^{\circ}\) and the length of the hypotenuse \(AB = 10\) units. We want to find the lengths of the other two sides \(BC\) (opposite to angle \(A\)) and \(AC\) (adjacent to angle \(A\)).

Solving the Designed Problem

Step 1: Recall the trigonometric ratios for a right - triangle

In a right - triangle, the sine of an angle \(\theta\) is defined as \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and the cosine of an angle \(\theta\) is defined as \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\).

Step 2: Find the length of \(BC\) (opposite to \(\angle A\))

We know that \(\theta = 30^{\circ}\), the hypotenuse \(AB = 10\), and we want to find \(BC\) (opposite side). Using the sine ratio:
\(\sin A=\frac{BC}{AB}\)
Substitute \(A = 30^{\circ}\) and \(AB = 10\) into the formula:
\(\sin30^{\circ}=\frac{BC}{10}\)
Since \(\sin30^{\circ}=\frac{1}{2}\), we can solve for \(BC\):
\(BC = 10\times\sin30^{\circ}=10\times\frac{1}{2} = 5\) units.

Step 3: Find the length of \(AC\) (adjacent to \(\angle A\))

Using the cosine ratio: \(\cos A=\frac{AC}{AB}\)
Substitute \(A = 30^{\circ}\) and \(AB = 10\) into the formula:
\(\cos30^{\circ}=\frac{AC}{10}\)
We know that \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), so:
\(AC = 10\times\cos30^{\circ}=10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\approx5\times1.732 = 8.66\) units.

Self - Assessment (Using the Rubric)

• Problem Design: The problem is clear. It provides a right - triangle with a known angle (other than the right angle) and the hypotenuse, and asks for the other two sides. This is a valid trigonometry problem as it uses the sine and cosine ratios.
• Solution: The solution correctly applies the trigonometric ratios (sine and cosine) for a right - triangle. The substitution of values and the calculation of the side lengths are correct. For example, using \(\sin30^{\circ}=\frac{1}{2}\) to find the opposite side and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\) to find the adjacent side is in line with the trigonometric principles for right - triangles.

Final Answer

The length of \(BC\) is \(5\) units and the length of \(AC\) is \(5\sqrt{3}\approx8.66\) units.

Answer:

Designing the Trigonometry Problem

Let's consider a right - triangle problem. Suppose we have a right - triangle \(ABC\) with a right angle at \(C\). We know that angle \(A = 30^{\circ}\) and the length of the hypotenuse \(AB = 10\) units. We want to find the lengths of the other two sides \(BC\) (opposite to angle \(A\)) and \(AC\) (adjacent to angle \(A\)).

Solving the Designed Problem

Step 1: Recall the trigonometric ratios for a right - triangle

In a right - triangle, the sine of an angle \(\theta\) is defined as \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and the cosine of an angle \(\theta\) is defined as \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\).

Step 2: Find the length of \(BC\) (opposite to \(\angle A\))

We know that \(\theta = 30^{\circ}\), the hypotenuse \(AB = 10\), and we want to find \(BC\) (opposite side). Using the sine ratio:
\(\sin A=\frac{BC}{AB}\)
Substitute \(A = 30^{\circ}\) and \(AB = 10\) into the formula:
\(\sin30^{\circ}=\frac{BC}{10}\)
Since \(\sin30^{\circ}=\frac{1}{2}\), we can solve for \(BC\):
\(BC = 10\times\sin30^{\circ}=10\times\frac{1}{2} = 5\) units.

Step 3: Find the length of \(AC\) (adjacent to \(\angle A\))

Using the cosine ratio: \(\cos A=\frac{AC}{AB}\)
Substitute \(A = 30^{\circ}\) and \(AB = 10\) into the formula:
\(\cos30^{\circ}=\frac{AC}{10}\)
We know that \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), so:
\(AC = 10\times\cos30^{\circ}=10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\approx5\times1.732 = 8.66\) units.

Self - Assessment (Using the Rubric)

• Problem Design: The problem is clear. It provides a right - triangle with a known angle (other than the right angle) and the hypotenuse, and asks for the other two sides. This is a valid trigonometry problem as it uses the sine and cosine ratios.
• Solution: The solution correctly applies the trigonometric ratios (sine and cosine) for a right - triangle. The substitution of values and the calculation of the side lengths are correct. For example, using \(\sin30^{\circ}=\frac{1}{2}\) to find the opposite side and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\) to find the adjacent side is in line with the trigonometric principles for right - triangles.

Final Answer

The length of \(BC\) is \(5\) units and the length of \(AC\) is \(5\sqrt{3}\approx8.66\) units.