Calculating Trigonometric Products in a Right-Angled Triangle

We are given a right-angled triangle ABC, with the right angle located at vertex B. The lengths of the two sides forming the right angle are provided: AB = 12 cm and BC = 5 cm.

To find the values of trigonometric ratios like sine and cosine, we first need to determine the length of the hypotenuse, AC. We can use the Pythagorean theorem for this.

Applying the Pythagorean Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. For triangle ABC, where ∠B = 90°:

\(AC^2 = AB^2 + BC^2\)

Substituting the given values:

\(AC^2 = (12\, \text{cm})^2 + (5\, \text{cm})^2\)

\(AC^2 = 144\, \text{cm}^2 + 25\, \text{cm}^2\)

\(AC^2 = 169\, \text{cm}^2\)

To find AC, we take the square root of 169:

\(AC = \sqrt{169}\, \text{cm}\)

\(AC = 13\, \text{cm}\)

So, the length of the hypotenuse is 13 cm.

Calculating sin A and cos A

For angle A:

• The side opposite to angle A is BC.
• The side adjacent to angle A is AB.
• The hypotenuse is AC.

The trigonometric ratios are defined as:

\(\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5}{13}\)

\(\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12}{13}\)

Calculating sin A cos A

Now, we find the product of sin A and cos A:

\(\sin A \cos A = \left(\frac{5}{13}\right) \times \left(\frac{12}{13}\right)\)

\(\sin A \cos A = \frac{5 \times 12}{13 \times 13}\)

\(\sin A \cos A = \frac{60}{169}\)

Calculating sin C and cos C

For angle C:

• The side opposite to angle C is AB.
• The side adjacent to angle C is BC.
• The hypotenuse is AC.

The trigonometric ratios for angle C are:

\(\sin C = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{12}{13}\)

\(\cos C = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{5}{13}\)

Calculating sin C cos C

Now, we find the product of sin C and cos C:

\(\sin C \cos C = \left(\frac{12}{13}\right) \times \left(\frac{5}{13}\right)\)

\(\sin C \cos C = \frac{12 \times 5}{13 \times 13}\)

\(\sin C \cos C = \frac{60}{169}\)

Summary of Results

We have calculated the required values:

• sin A cos A = \(\frac{60}{169}\)
• sin C cos C = \(\frac{60}{169}\)

Therefore, the values of sin A cos A and sin C cos C are \(\frac{60}{169}\) and \(\frac{60}{169}\) respectively.

Revision Table: Right Triangle Trigonometry

Additional Information: Complementary Angles

In a right-angled triangle ABC with ∠B = 90°, angles A and C are complementary, meaning A + C = 90°. This leads to interesting relationships between their trigonometric ratios:

• sin A = cos C
• cos A = sin C
• tan A = cot C

In this problem, we found sin A = 5/13 and cos C = 5/13, confirming sin A = cos C. Similarly, cos A = 12/13 and sin C = 12/13, confirming cos A = sin C.

This property is why sin A cos A = (5/13) * (12/13) and sin C cos C = (12/13) * (5/13) result in the same value (60/169). The product of sin and cos of two complementary angles derived from the sides will always be the same value if the sides are the same length.