Flagpole Distance D664De
1. **Problem statement:** Jack and Sangita stand on opposite sides of a 10-m flagpole. Jack sees the top of the pole at an angle of elevation of 50°, and Sangita sees it at 35°. We need to find the distance between Jack and Sangita.
2. **Understanding the problem:** The flagpole is vertical and 10 m tall. Jack and Sangita are on opposite sides, so the flagpole forms a triangle with their positions.
3. **Key insight:** We can model the situation as a triangle where the flagpole is the height, and the distances from Jack and Sangita to the base of the pole are the adjacent sides to the angles of elevation.
4. **Formulas:** For each person, the distance to the pole base can be found using the tangent function:
$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$
where opposite = 10 m (height of pole), adjacent = distance from person to pole base.
5. **Calculate distances:**
- For Jack:
$$d_J = \frac{10}{\tan(50^\circ)}$$
- For Sangita:
$$d_S = \frac{10}{\tan(35^\circ)}$$
6. **Calculate values:**
- $\tan(50^\circ) \approx 1.1918$
- $\tan(35^\circ) \approx 0.7002$
So,
$$d_J = \frac{10}{1.1918} \approx 8.39\,m$$
$$d_S = \frac{10}{0.7002} \approx 14.28\,m$$
7. **Find total distance:** Since Jack and Sangita are on opposite sides of the pole, the total distance between them is:
$$d = d_J + d_S = 8.39 + 14.28 = 22.67\,m$$
8. **Answer:** Jack and Sangita are approximately 22.67 meters apart.
9. **Regarding the law to use:** You do not need the sine or cosine law here. Using the tangent function with right triangles formed by the pole and the line of sight is sufficient.