Points A and B 1000 meter apart are plotted on a straight highway running east and west. from A, the bearing of tower C is N32*W and from be, the bearing of C is N64*E. approximate the shortest distance of the tower from the highway.

please explain........

Ikaw dennis, wala kang halaga sa mundo.

dennis yung titi mo , supot ... hehe

Bakit wala yung sagot? sure na yan?

373.81 m

To approximate the shortest distance of tower C from the highway, we can use trigonometry and the given information of the bearings.

1. Draw a diagram: Start by drawing a straight east-west highway line and label two points A and B, 1000 meters apart.

2. Determine the positions: Based on the given information, point A has a bearing of N32°W to tower C, and point B has a bearing of N64°E to tower C. This means that tower C is located somewhere between points A and B.

3. Find the intersection point: To find the position of tower C, draw two lines from points A and B representing their respective bearings. Extend these lines until they intersect. The intersection point is the approximate location of tower C.

4. Measure the distance: Measure the perpendicular distance from the intersection point to the highway. This distance represents the shortest distance of the tower from the highway.

By following these steps and using basic trigonometry, you can approximate the shortest distance of tower C from the highway.

The problem can be represented by a triangle with a base equal to 1000 m and

base angles A and B. The 3rd angle is C.

A = 90o-32o = 58o.
B = 90o-64o = 26o.
C = 180-58-26 = 96o.

c/sinC = a/sinA.
1000/sin96 = a/sin58
Multiply bothsides by sin58:
a = 1000*sin58/sin96 = 853 m.

b/sinB = c/sinC.
b/sin26 = 1000/sin96
b = 1000*sin26/sin96 = 441 m.

Side b represents the shortest dist.
from highway to Tower C.

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