In the figure, a man stands at a point P which is 100 m due west of a
building OT and finds that the angle of elevation of T is 40 degree. He
then walks 140 m to another point Q and finds that the angle of elevation
of T is 55 degree. O, P and Q lie on the same horizontal ground.
a) Find the height ( OT ) of the building.
b) Find the distance between O and Q.
c) Find the compass bearing of Q from O.
Figure: http://img704.imageshack.us/img704/4104/maths30.jpg
F4 Maths 3-D problem - -
2010-03-22 2:20 am
回答 (1)
2010-03-22 4:55 am
✔ 最佳答案
a)In ΔTOP:
tan40º = OT/(100 m)
OT = 100 tan40º m
OT = 83.9 m
Height of the building = 83.9 m
b)
In ΔTOQ:
tan55º = (83.9 m)/OQ
OQ = 83.9/tan55º m
OQ = 58.7 m
Distance between O and Q = 58.7 m
c)
By cosine law:
cos∠POQ = (OP² + OQ² - PQ²) / (2 * OP * OQ)
cos∠POQ = (100² + 58.7² - 140²) / (2 * 100 * 58.7)
∠POQ = 121.6º
(270 - 121.6)º = 148.4º
Compass bearing of Q from O = 148.4º
(equivalent to S 31.6º N)
參考: micatkie
收錄日期: 2021-04-13 17:09:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100321000051KK01386