We are given a right triangle with sides $a = 77.1$ yd and $b = 42.1$ yd. Let's solve for the hypotenuse $c$, and the angles $A$ and $B$.

Step 1: Solve for the hypotenuse $c$

The Pythagorean theorem relates the sides of a right triangle: $c = \sqrt{a^2 + b^2}$ Substitute $a = 77.1$ yd and $b = 42.1$ yd: $c = \sqrt{(77.1)^2 + (42.1)^2}$ $c = \sqrt{5944.41 + 1772.41} = \sqrt{7716.82}$ $c \approx 87.8 \, \text{yd}$

Step 2: Solve for angle $A$

We can use the tangent function to find angle $A$ since we know the opposite and adjacent sides: $\tan(A) = \frac{b}{a}$ $\tan(A) = \frac{42.1}{77.1} \approx 0.546$ Now, use the arctangent (inverse tangent) to find $A$: $A = \tan^{-1}(0.546) \approx 28.6^\circ$ Convert the decimal part to minutes by multiplying by 60: $0.6 \times 60 = 36 \, \text{minutes}$ So, $A \approx 28^\circ 36'$.

Step 3: Solve for angle $B$

Since the sum of angles in a triangle is $180^\circ$ and one angle is $90^\circ$ (right triangle): $B = 90^\circ - A$ $B = 90^\circ - 28^\circ 36' = 61^\circ 24'$

Final Answers:

• $c \approx 87.8$ yd
• $A \approx 28^\circ 36'$
• $B \approx 61^\circ 24'$

Would you like more details on any part of the solution?

Here are five related questions to expand on this:

1. How would the solution change if we swapped sides $a$ and $b$?
2. What is the area of the triangle given the sides $a$ and $b$?
3. Can we solve for the angles using sine or cosine instead of tangent?
4. How would we solve if only the hypotenuse $c$ and one side $a$ were given?
5. What are the applications of trigonometry in real-world problems?

Tip: Always keep your calculator in degree mode when working with angles in degrees!