Dado el triangulo rectanyb \( A B C \), donde \( K C=90^{\circ} \) \( \nleftarrow B=72 \) y \( C=38 \mathrm{~m} \) hallay elementos faltantes

Alright, I need to solve the problem involving triangle \( ABC \) where \( \angle B = 72^\circ \) and \( BC = 38 \) meters. The triangle is right-angled at \( C \), so \( \angle C = 90^\circ \). I need to find the missing elements of the triangle. First, let's recall the properties of a right-angled triangle. In a right-angled triangle, the sum of the other two angles is \( 90^\circ \). Given that \( \angle B = 72^\circ \), I can find \( \angle A \) by subtracting \( 72^\circ \) from \( 90^\circ \). So, \( \angle A = 90^\circ - 72^\circ = 18^\circ \). Now, I have the measures of all three angles: \( \angle A = 18^\circ \), \( \angle B = 72^\circ \), and \( \angle C = 90^\circ \). The side opposite the right angle is \( BC = 38 \) meters. I need to find the lengths of the other two sides, \( AB \) and \( AC \). To find \( AB \) and \( AC \), I can use trigonometric ratios. Since \( \angle C \) is \( 90^\circ \), \( BC \) is the hypotenuse of the triangle. The side opposite \( \angle A \) is \( BC \), and the side opposite \( \angle B \) is \( AC \). Using the sine function for \( \angle A \): \[ \sin(\angle A) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{BC} \] \[ \sin(18^\circ) = \frac{AC}{38} \] \[ AC = 38 \times \sin(18^\circ) \] Calculating \( \sin(18^\circ) \): \[ \sin(18^\circ) \approx 0.3090 \] \[ AC \approx 38 \times 0.3090 \approx 11.742 \text{ meters} \] Next, using the cosine function for \( \angle A \): \[ \cos(\angle A) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{BC} \] \[ \cos(18^\circ) = \frac{AB}{38} \] \[ AB = 38 \times \cos(18^\circ) \] Calculating \( \cos(18^\circ) \): \[ \cos(18^\circ) \approx 0.9511 \] \[ AB \approx 38 \times 0.9511 \approx 36.1418 \text{ meters} \] So, the missing elements are: - \( AC \approx 11.742 \) meters - \( AB \approx 36.1418 \) meters But wait, let me double-check these calculations to ensure accuracy. Using a calculator for precise values: \[ \sin(18^\circ) \approx 0.309016994 \] \[ AC = 38 \times 0.309016994 \approx 11.742645772 \text{ meters} \] \[ \cos(18^\circ) \approx 0.951056516 \] \[ AB = 38 \times 0.951056516 \approx 36.141147608 \text{ meters} \] Rounding to two decimal places: \[ AC \approx 11.74 \text{ meters} \] \[ AB \approx 36.14 \text{ meters} \] Everything checks out. Therefore, the complete set of elements for triangle \( ABC \) is: - \( \angle A = 18^\circ \) - \( \angle B = 72^\circ \) - \( \angle C = 90^\circ \) - \( AB \approx 36.14 \) meters - \( AC \approx 11.74 \) meters - \( BC = 38 \) meters