The image shows a right triangle formed by a building, the ground, and a diagonal line. The given information includes:

• One angle of the triangle is 45°.
• The hypotenuse (the diagonal line) measures 50 meters.
• The building's height is labeled as $a$, the base is labeled $b$, and the angle between the building and the hypotenuse is labeled $\beta$.

To solve for $a$, $\beta$, and $b$:

Step 1: Use the angle $45^\circ$

Since one of the angles is 45° and it's a right triangle, we can use trigonometric ratios. Let's recall the definitions:

• $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$
• $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$
• $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

Part a: Find $a$

We can use the sine function since $a$ is the opposite side to the 45° angle and the hypotenuse is 50 m:

$\sin(45^\circ) = \frac{a}{50}$

Since $\sin(45^\circ) = \frac{\sqrt{2}}{2}$:

$\frac{\sqrt{2}}{2} = \frac{a}{50}$

Solving for $a$:

$a = 50 \times \frac{\sqrt{2}}{2} = 50 \times 0.707 \approx 35.35 \, \text{meters}$

Part b: Find $\beta$

Since the angles in a triangle sum up to 180°, and we already know that one angle is 45° and another is 90°:

$\beta = 180^\circ - 90^\circ - 45^\circ = 45^\circ$

Part c: Find $b$

We can use the cosine function to find $b$, which is adjacent to the 45° angle:

$\cos(45^\circ) = \frac{b}{50}$

Since $\cos(45^\circ) = \frac{\sqrt{2}}{2}$:

$\frac{\sqrt{2}}{2} = \frac{b}{50}$

Solving for $b$:

$b = 50 \times \frac{\sqrt{2}}{2} = 50 \times 0.707 \approx 35.35 \, \text{meters}$

Summary of answers:

• $a \approx 35.35 \, \text{m}$
• $\beta = 45^\circ$
• $b \approx 35.35 \, \text{m}$

Let me know if you want more details or further clarifications.

Here are 5 related questions to expand on this:

1. How would the values change if the hypotenuse was different?
2. How do trigonometric functions apply in solving real-world problems?
3. What is the relationship between the sine and cosine of complementary angles?
4. Can you solve for the area of the triangle given the current values?
5. How would the problem change if the angle $\beta$ were different?

Tip: Remember that in right triangles, complementary angles always add up to 90°, and their sine and cosine functions are related.