Diane and Limore are looking at their school from the area beside the gym. From ambers point of view, the top of the school is at an angle of elevation of 48. From joeys point of view, directly closer to the school it is 68 degrees. the school is 30m high. How far apart are the two students?
Round to nearest tenth of a metere.
I keep getting the wrong answer my answer was 13m but it says it is 14.9 meters?
from amber
tan 48 = 30/d
d = 30 / tan 48 = 30/1.11 = 27 meters total to building
from joey
e = 30/tan 68 = 30 / 2.48 = 12.1 from joey to building
27 - 12.1 = 14.9
I agree with your book or whatever
To find the distance between the two students, we can use trigonometry and the concept of similar triangles. Here's how you can solve the problem step by step:
1. Draw a diagram to visualize the situation. Label the school's height (30m) and the angles of elevation for Amber (48 degrees) and Joey (68 degrees).
2. Note that the triangles formed by the students and the school are similar, as they have the same corresponding angles. This allows us to set up a proportion.
3. Let's assume the distance between the students is "d" meters. The proportion can be set up as follows:
d / (d + 30) = tan(48 degrees) / tan(68 degrees)
Here, we are using the tangent function (tan) because the angles of elevation are given.
4. Now, solve this proportion for "d." Start by calculating the tangent of each angle:
tan(48 degrees) ≈ 1.1106
tan(68 degrees) ≈ 2.3387
5. Substitute these values back into the proportion:
d / (d + 30) = 1.1106 / 2.3387
6. Cross-multiply and solve for "d":
2.3387d = 1.1106(d + 30)
2.3387d = 1.1106d + 33.318
2.3387d - 1.1106d = 33.318
1.2281d = 33.318
d ≈ 27.119
7. Round the distance to the nearest tenth:
d ≈ 27.1 meters (rounded to the nearest tenth)
Therefore, the two students, Diane and Limore, are approximately 27.1 meters apart. It seems there was a rounding error in your calculations, leading to the incorrect answer of 13 meters.