Let's solve each triangle step by step.

Problem 1

We have a right triangle $\triangle PQR$, where:

• $p = 19 \, \text{cm}$ (adjacent side),
• $\angle PQR = 37^\circ$,
• $\angle PRQ = 90^\circ$,
• We need to find $q$ (opposite side), $r$ (hypotenuse), and the missing angle $\angle QPR$.

1. Find the hypotenuse $r$: Use the cosine function: $\cos(37^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{p}{r}$ $r = \frac{p}{\cos(37^\circ)} = \frac{19}{\cos(37^\circ)}$ Using $\cos(37^\circ) \approx 0.7986$: $r \approx \frac{19}{0.7986} \approx 24 \, \text{cm}$

2. Find the opposite side $q$: Use the tangent function: $\tan(37^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{q}{p}$ $q = p \cdot \tan(37^\circ) = 19 \cdot \tan(37^\circ)$ Using $\tan(37^\circ) \approx 0.7536$: $q \approx 19 \cdot 0.7536 \approx 14 \, \text{cm}$

3. Find the missing angle $\angle QPR$: In a triangle, the sum of angles is $180^\circ$. Since one angle is $90^\circ$: $\angle QPR = 90^\circ - 37^\circ = 53^\circ$

Final Answer for Problem 1:

• $r \approx 24 \, \text{cm}$,
• $q \approx 14 \, \text{cm}$,
• $\angle QPR = 53^\circ$.

Problem 2

We have a right triangle $\triangle MNL$, where:

• $m = 37 \, \text{cm}$ (hypotenuse),
• $\angle MNL = 51^\circ$,
• $\angle MLN = 90^\circ$,
• We need to find $n$ (adjacent side), $c$ (opposite side), and the missing angle $\angle NML$.

1. Find the adjacent side $n$: Use the cosine function: $\cos(51^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{n}{m}$ $n = m \cdot \cos(51^\circ) = 37 \cdot \cos(51^\circ)$ Using $\cos(51^\circ) \approx 0.6293$: $n \approx 37 \cdot 0.6293 \approx 23 \, \text{cm}$

2. Find the opposite side $c$: Use the sine function: $\sin(51^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{c}{m}$ $c = m \cdot \sin(51^\circ) = 37 \cdot \sin(51^\circ)$ Using $\sin(51^\circ) \approx 0.7771$: $c \approx 37 \cdot 0.7771 \approx 29 \, \text{cm}$

3. Find the missing angle $\angle NML$: In a triangle, the sum of angles is $180^\circ$. Since one angle is $90^\circ$: $\angle NML = 90^\circ - 51^\circ = 39^\circ$

Final Answer for Problem 2:

• $n \approx 23 \, \text{cm}$,
• $c \approx 29 \, \text{cm}$,
• $\angle NML = 39^\circ$.

Summary:

Problem 1:

• $r \approx 24 \, \text{cm}, q \approx 14 \, \text{cm}, \angle QPR = 53^\circ$.

Problem 2:

• $n \approx 23 \, \text{cm}, c \approx 29 \, \text{cm}, \angle NML = 39^\circ$.

Would you like a detailed breakdown for any step?

Follow-up Questions:

1. How can you use trigonometric functions to solve any right triangle?
2. What is the significance of the complementary angles in a right triangle?
3. How would the solution change if the given angle were obtuse?
4. Can the Pythagorean theorem be used to verify these calculations?
5. What are other real-life applications of solving right triangles?

Tip:

Always double-check your trigonometric ratios and angle measures when solving triangles to avoid rounding errors!