Analyzing the problem:

1. First image: The problem involves determining how far you have to travel horizontally along your street before turning north to reach your friend’s house. This involves solving a right triangle:

   • Hypotenuse ($c$) = 10.0 km
   • Vertical leg ($b$) = 6.0 km
   • Unknown horizontal leg ($a$).

   Using the Pythagorean theorem: $c^2 = a^2 + b^2$

   Rearrange to find $a$: $a = \sqrt{c^2 - b^2}$

2. Second image: It asks for the sine of the angle $\theta$ at the friend’s house:

   • $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{c}$.

Solutions:

1. Horizontal distance $a$: Substituting values into the Pythagorean theorem: $a = \sqrt{10.0^2 - 6.0^2} = \sqrt{100 - 36} = \sqrt{64} = 8.0 \, \text{km}.$ So, the distance is 8.0 km.

2. Sine of angle $\theta$: Using the sine definition: $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6.0}{10.0} = 0.6.$

Final Answers:

1. Horizontal distance: $8.0 \, \text{km}$.
2. Sine of angle $\theta$: $0.6$.

Would you like me to go into further detail or explore related concepts?

5 Related Questions:

1. How can the Pythagorean theorem be used in real-life navigation?
2. What is the relationship between sine, cosine, and tangent in right triangles?
3. How do you calculate the angle $\theta$ using sine or other trigonometric ratios?
4. What are the different applications of trigonometry in GPS and mapping?
5. How do you use the Pythagorean theorem to solve for a hypotenuse?

Tip:

Remember, trigonometric ratios (sine, cosine, tangent) are very useful tools for understanding angles and distances in navigation and real-world problems!