Question 12 (9 points)

Question

Listen

C

12 cm

B

9cm

A

Using the triangle above, answer the following questions:

Find the length of the missing side (CB) using Pythagorean's theorem. (2 marks).

Calculate the angle 'A' using trigonometric functions. (2 marks)

Using angle 'A' as your reference point, label all three sides of this triangle using

the terms hypotenuse, adjacent and opposite. (3 marks)

Using angle 'A' as your reference point, provide the trigonometric ratios for the following: (2 marks)
tan 0 =
sin 0 =

Answer 1

To solve the problems, we can start with the information given about the triangle, which we will label with the following vertices:

• \( A \) (angle \( A \) is at vertex \( A \))
• \( B \) (the side opposite angle \( A \) which is 9 cm)
• \( C \) (the side opposite vertex \( B \) which is 12 cm)

1. Finding the length of the missing side (CB) using the Pythagorean theorem.

To find the length of side \( CB \), we will use the Pythagorean theorem, which states:

\[ a^2 + b^2 = c^2 \]

where \( c \) is the hypotenuse (the longest side), and \( a \) and \( b \) are the legs of the triangle. In this triangle:

• \( a = AB = 9 , \text{cm} \) (one leg)
• \( b = CB \) (the other leg, which we need to find)
• \( c = AC = 12 , \text{cm} \) (the hypotenuse)

Substituting the known values into the formula:

\[ 9^2 + CB^2 = 12^2 \]

Calculating those squares:

\[ 81 + CB^2 = 144 \]

Now, isolate \( CB^2 \):

\[ CB^2 = 144 - 81 \] \[ CB^2 = 63 \]

Taking the square root:

\[ CB = \sqrt{63} \approx 7.94 , \text{cm} \]

2. Calculating the angle 'A' using trigonometric functions.

To find angle \( A \), we can use the sine, cosine, or tangent functions. We will calculate it using the tangent function:

\[ \tan(A) = \frac{\text{opposite}}{\text{adjacent}} \]

Here, the opposite side is \( AB = 9 , \text{cm} \) and the adjacent side is \( CB \approx 7.94 , \text{cm} \):

\[ \tan(A) = \frac{9}{7.94} \]

To find \( A \):

\[ A = \tan^{-1}\left(\frac{9}{7.94}\right) \] Calculating:

\[ A \approx \tan^{-1}(1.13) \approx 48.37^\circ \]

3. Labeling the sides of the triangle.

Using angle \( A \) as the reference point:

• Hypotenuse: \( AC = 12 , \text{cm} \)
• Adjacent: \( CB \approx 7.94 , \text{cm} \) (the side next to angle \( A \))
• Opposite: \( AB = 9 , \text{cm} \) (the side opposite angle \( A \))

4. Providing the trigonometric ratios.

Using angle \( A \):

• Tangent ratio: \[ \tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{CB} = \frac{9}{7.94} \]

• Sine ratio: \[ \sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{9}{12} \]

\[ \sin(A) = 0.75 \]

Summary of results:

1. \( CB \approx 7.94 , \text{cm} \)
2. \( A \approx 48.37^\circ \)
3. Sides labeled:
   • Hypotenuse: \( AC = 12 , \text{cm} \)
   • Adjacent: \( CB \approx 7.94 , \text{cm} \)
   • Opposite: \( AB = 9 , \text{cm} \)
4. Trigonometric ratios:
   • \( \tan(A) = \frac{9}{7.94} \)
   • \( \sin(A) = \frac{9}{12} = 0.75 \)