Let's solve each triangle step-by-step, providing all necessary calculations and explanations. We'll use trigonometric relationships for right triangles (sine, cosine, tangent) and the Pythagorean theorem. --- ## **Problem 1** Given: - \( A = 30.75^\circ \) - \( b = 2' 3\frac{5}{8}" \) (convert this to inches first) - \( C = 90^\circ \) Let's denote: - \( a \): side opposite angle \( A \) - \( b \): side adjacent to angle \( A \) (given) - \( c \): hypotenuse ### **Step 1: Convert \( b \) to inches** \( 2' = 24" \) So, \( b = 24" + 3\frac{5}{8}" = 24" + 3.625" = 27.625" \) ### **Step 2: Find angle \( B \)** Sum of angles in a triangle = \( 180^\circ \) \[ B = 180^\circ - 90^\circ - 30.75^\circ = 59.25^\circ \] --- ### **Step 3: Use Trigonometric Ratios** \[ \cos(A) = \frac{b}{c} \implies c = \frac{b}{\cos(A)} \] \[ \sin(A) = \frac{a}{c} \implies a = c \cdot \sin(A) \] Or: \[ \tan(A) = \frac{a}{b} \implies a = b \cdot \tan(A) \] --- ### **Step 4: Calculate \( a \) and \( c \)** #### **a) Find \( a \):** \[ a = b \cdot \tan(A) = 27.625 \times \tan(30.75^\circ) \] \[ \tan(30.75^\circ) \approx 0.5949 \] \[ a \approx 27.625 \times 0.5949 \approx 16.42" \] #### **b) Find \( c \):** \[ c = \frac{b}{\cos(A)} = \frac{27.625}{\cos(30.75^\circ)} \] \[ \cos(30.75^\circ) \approx 0.8587 \] \[ c \approx \frac{27.625}{0.8587} \approx 32.17" \] --- ### **Step 5: Convert answers to feet, inches, and fractions** #### **a) For \( a \approx 16.42" \):** - \( 16" \) and \( 0.42" \) - \( 0.42 \times 16 \approx 6.7 \) → \( \frac{7}{16}" \) (rounded to nearest \( 1/16 \)) - \( a \approx 1' 4\frac{7}{16}" \) #### **b) For \( c \approx 32.17" \):** - \( 32.17" = 2' 8.17" \) - \( 0.17 \times 16 \approx 2.7 \) → \( \frac{3}{16}" \) - \( c \approx 2' 8\frac{3}{16}" \) --- ### **Final Answers for Problem 1:** - \( \boxed{A = 30.75^\circ} \) - \( \boxed{B = 59.25^\circ} \) - \( \boxed{C = 90^\circ} \) - \( \boxed{a = 1' 4\frac{7}{16}''} \) - \( \boxed{b = 2' 3\frac{5}{8}''} \) (given) - \( \boxed{c = 2' 8\frac{3}{16}''} \) --- ## **Problem 2** Given: - \( a = 1' 6 \frac{7}{16}'' \) - \( c = 3' 5 \frac{1}{4}'' \) - \( C = 90^\circ \) ### **Step 1: Convert to inches** - \( a = 12" + 6.4375" = 18.4375" \) - \( c = 36" + 5.25" = 41.25" \) ### **Step 2: Find angles \( A \) and \( B \)** #### **a) Find \( A \):** \[ \sin(A) = \frac{a}{c} \] \[ A = \arcsin\left(\frac{18.4375}{41.25}\right) \] \[ \frac{18.4375}{41.25} \approx 0.4469 \] \[ A \approx \arcsin(0.4469) \approx 26.6^\circ \] #### **b) Find \( B \):** \[ B = 180^\circ - 90^\circ - 26.6^\circ = 63.4^\circ \] --- ### **Step 3: Find the missing side \( b \):** \[ b^2 = c^2 - a^2 \] \[ b = \sqrt{41.25^2 - 18.4375^2} = \sqrt{1702.56 - 340.03} = \sqrt{1362.53} \approx 36.92" \] #### **Convert \( b \) to feet, inches, and fractions:** - \( 36.92" = 3' 0.92" \) - \( 0.92 \times 16 = 14.7 \approx 15/16 \) - \( b \approx 3' 15/16'' \) --- ### **Final Answers for Problem 2:** - \( \boxed{A = 26.6^\circ} \) - \( \boxed{B = 63.4^\circ} \) - \( \boxed{C = 90^\circ} \) - \( \boxed{a = 1' 6\frac{7}{16}''} \) (given) - \( \boxed{b = 3' \frac{15}{16}''} \) - \( \boxed{c = 3' 5\frac{1}{4}''} \) (given) --- **If you need any intermediate steps explained in more detail, or want me to check the rounding, let me know!**