We are asked to find the missing elements (sides and angles), the perimeter, and the area of the given right triangles. a) We have a right triangle with side $a = 45 m$ and angle $C = 25^\circ$. We need to find side $b$, side $c$, angle $B$, the perimeter, and the area. b) We have a right triangle with sides $5 km$ and $12 km$. We need to find the hypotenuse $x$, and then calculate the perimeter and the area.

1. Problem Description

We are asked to find the missing elements (sides and angles), the perimeter, and the area of the given right triangles.

a) We have a right triangle with side $a = 45 m$ and angle $C = 25^\circ$. We need to find side $b$, side $c$, angle $B$, the perimeter, and the area.

b) We have a right triangle with sides $5 km$ and $12 km$. We need to find the hypotenuse $x$, and then calculate the perimeter and the area.

2. Solution Steps

a)

Since the sum of angles in a triangle is $180^\circ$ and we have a right triangle, we know that $A = 90^\circ$. Thus, $B = 180^\circ - 90^\circ - 25^\circ = 65^\circ$.

To find $b$, we use the sine function:

$sin(C) = \frac{c}{a}$

Therefore $c = a * sin(C)$. Then $c = 45 * sin(25^\circ) = 45 * 0.4226 \approx 19.02 m$

To find $b$, we use the cosine function:

$cos(C) = \frac{b}{a}$

Therefore $b = a * cos(C)$. Then $b = 45 * cos(25^\circ) = 45 * 0.9063 \approx 40.78 m$

Perimeter $P = a + b + c = 45 + 40.78 + 19.02 = 104.8 m$

Area $A = \frac{1}{2} * b * c = \frac{1}{2} * 40.78 * 19.02 \approx 387.83 m^2$

b)

We use the Pythagorean theorem to find the hypotenuse $x$:

$x^2 = 5^2 + 12^2 = 25 + 144 = 169$

$x = \sqrt{169} = 13 km$

Perimeter $P = 5 + 12 + 13 = 30 km$

Area $A = \frac{1}{2} * 5 * 12 = 30 km^2$

3. Final Answer

a)

$B = 65^\circ$

$c \approx 19.02 m$

$b \approx 40.78 m$

$P \approx 104.8 m$

$A \approx 387.83 m^2$

b)

$x = 13 km$

$P = 30 km$

$A = 30 km^2$