Understanding the Right-Angled Triangle

We are given a right-angled triangle where the angle at vertex B is 90 degrees. The lengths of the sides are 6, 8, and 10 units.

The problem states that vertex C is opposite the side with length 8 units. In a triangle, the side opposite a vertex is the side that does not connect to that vertex. Therefore, the side opposite vertex C is side AB. This means the length of side AB is 8 units.

Since the triangle is right-angled at B, the side AC is the hypotenuse. The given hypotenuse length is 10 units, so AC = 10.

The remaining side is BC. Its length must be 6 units. We can confirm this using the Pythagorean theorem ($a^2 + b^2 = c^2$):

Check: $AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$.

And the hypotenuse squared: $AC^2 = 10^2 = 100$.

Since $100 = 100$, the side lengths fit the Pythagorean theorem. The sides of our triangle are AB = 8, BC = 6, and AC = 10.

Calculating Trigonometric Values tan A and cos C

To find the value of $\tan^2 A + \cos^2 C$, we first need to determine the values of $\tan A$ and $\cos C$. We need to identify the opposite, adjacent, and hypotenuse sides relative to angles A and C.

For angle A:

• Opposite side = BC = 6
• Adjacent side = AB = 8
• Hypotenuse = AC = 10

For angle C:

• Opposite side = AB = 8
• Adjacent side = BC = 6
• Hypotenuse = AC = 10

Recall the definitions of the trigonometric ratios in a right-angled triangle:

• Tangent of an angle ($\tan$): $\frac{\text{Opposite}}{\text{Adjacent}}$
• Cosine of an angle ($\cos$): $\frac{\text{Adjacent}}{\text{Hypotenuse}}$

Calculating tan A

Using the definition for $\tan A$ and the side lengths:

$\tan A = \frac{BC}{AB} = \frac{6}{8}$

Simplifying the fraction, we get:

$\tan A = \frac{3}{4}$

Calculating cos C

Using the definition for $\cos C$ and the side lengths:

$\cos C = \frac{BC}{AC} = \frac{6}{10}$

Simplifying the fraction, we get:

$\cos C = \frac{3}{5}$

Evaluating the Expression tan^2 A + cos^2 C

Now we need to square the values of $\tan A$ and $\cos C$ and then add them together.

Calculating tan^2 A

Squaring the value of $\tan A$:

$\tan^2 A = (\tan A)^2 = (\frac{3}{4})^2 = \frac{3^2}{4^2} = \frac{9}{16}$

Calculating cos^2 C

Squaring the value of $\cos C$:

$\cos^2 C = (\cos C)^2 = (\frac{3}{5})^2 = \frac{3^2}{5^2} = \frac{9}{25}$

Summing the Squared Values

Finally, we add $\tan^2 A$ and $\cos^2 C$:

$\tan^2 A + \cos^2 C = \frac{9}{16} + \frac{9}{25}$

To add these fractions, we find a common denominator. The least common multiple of 16 and 25 is $16 \times 25 = 400$.

Convert the fractions to have the denominator 400:

$\frac{9}{16} = \frac{9 \times 25}{16 \times 25} = \frac{225}{400}$

$\frac{9}{25} = \frac{9 \times 16}{25 \times 16} = \frac{144}{400}$

Now add the converted fractions:

$\frac{225}{400} + \frac{144}{400} = \frac{225 + 144}{400} = \frac{369}{400}$

Final Answer

The value of the expression $\tan^2 A + \cos^2 C$ is $\frac{369}{400}$.