To solve the right triangle $\triangle ABC$ where $C = 90^\circ$, side $a = 75.4$ yards, and side $b = 41.8$ yards, we need to find:

1. The length of side $c$ (the hypotenuse).
2. The angle $A$ in degrees and minutes.
3. The angle $B$ in degrees and minutes.

1. Find the hypotenuse $c$ using the Pythagorean theorem:

$c = \sqrt{a^2 + b^2}$ Substituting $a = 75.4$ yards and $b = 41.8$ yards: $c = \sqrt{75.4^2 + 41.8^2}$

Let me calculate this.

$c = \sqrt{75.4^2 + 41.8^2} = \sqrt{5680.36 + 1747.24} = \sqrt{7427.6}$ $c \approx 86.2 \text{ yards}$

2. Find the angle $A$:

To find angle $A$, we use the tangent function: $\tan(A) = \frac{a}{b} = \frac{75.4}{41.8}$ $A = \tan^{-1} \left( \frac{75.4}{41.8} \right)$

Let me calculate this.

$A \approx \tan^{-1}(1.8048) \approx 61.5^\circ$

Convert the decimal part into minutes: $0.5^\circ \times 60 = 30 \text{ minutes}$ Thus, $A \approx 61^\circ 30'$.

3. Find the angle $B$:

Since $\triangle ABC$ is a right triangle, the sum of angles is $180^\circ$. Therefore, angle $B$ can be calculated as: $B = 90^\circ - A = 90^\circ - 61^\circ 30'$ $B \approx 28^\circ 30'$

Final Answer:

• $c \approx 86.2$ yards
• $A \approx 61^\circ 30'$
• $B \approx 28^\circ 30'$

Would you like more details or have any questions?

Related Questions:

1. What is the sine of angle $A$?
2. What is the cosine of angle $B$?
3. How would you find the area of triangle $ABC$?
4. Can the triangle's area be calculated using trigonometry?
5. What is the ratio of the triangle's sides?

Tip: Always remember to round to the required precision after each calculation to avoid errors.