The problem asks us to find the values of $x$ and $y$ in two right triangles using trigonometric ratios. In the first triangle, the hypotenuse is 12cm, and one angle is 45 degrees. We need to find $x$ (opposite) and $y$ (adjacent). In the second triangle, the hypotenuse is 6km, one angle is 30 degrees, and we need to find $x$ (opposite) and $y$ (adjacent). Then, we're asked to use the values found to calculate another value of $y$ given $x = 3km$ and an angle of $30^\circ$.

1. Problem Description

The problem asks us to find the values of $x$ and $y$ in two right triangles using trigonometric ratios. In the first triangle, the hypotenuse is 12cm, and one angle is 45 degrees. We need to find $x$ (opposite) and $y$ (adjacent). In the second triangle, the hypotenuse is 6km, one angle is 30 degrees, and we need to find $x$ (opposite) and $y$ (adjacent). Then, we're asked to use the values found to calculate another value of $y$ given $x = 3km$ and an angle of $30^\circ$.

2. Solution Steps

For the first triangle:

(i) To find $y$ (adjacent), we can use the cosine function:

$cos(\theta) = \frac{adjacent}{hypotenuse}$

$cos(45^\circ) = \frac{y}{12}$

$y = 12 \cdot cos(45^\circ)$

$y = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2}$

$y \approx 8.485 cm$

(ii) To find $x$ (opposite), we can use the sine function:

$sin(\theta) = \frac{opposite}{hypotenuse}$

$sin(45^\circ) = \frac{x}{12}$

$x = 12 \cdot sin(45^\circ)$

$x = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2}$

$x \approx 8.485 cm$

For the second triangle:

(i) To find $x$ (opposite), we can use the sine function:

$sin(\theta) = \frac{opposite}{hypotenuse}$

$sin(30^\circ) = \frac{x}{6}$

$x = 6 \cdot sin(30^\circ)$

$x = 6 \cdot \frac{1}{2} = 3 km$

(ii) The final part involves using the tangent function and the previously calculated value of $x=3km$ and the given angle of $30^\circ$ to find $y$:

$tan(\theta) = \frac{opposite}{adjacent}$

$tan(30^\circ) = \frac{3}{y}$

$y = \frac{3}{tan(30^\circ)}$

$y = \frac{3}{\frac{1}{\sqrt{3}}} = 3\sqrt{3}$

$y \approx 5.196 km$

3. Final Answer

$y \approx 8.485 cm$

$x \approx 8.485 cm$

$x = 3 km$

$y \approx 5.196 km$