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

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

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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

En el triángulo rectángulo $$ ABC $$ con $$ \angle C = 90^\circ $$, $$ \angle B = 72^\circ $$, y $$ BC = 38 $$ metros, los elementos faltantes son: - $$ \angle A = 18^\circ $$ - $$ AB \approx 36.14 $$ metros - $$ AC \approx 11.74 $$ metros

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

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