Tuesday

September 23, 2014

September 23, 2014

Posted by **Anonymous** on Tuesday, February 5, 2013 at 8:33am.

- trig -
**Reiny**, Tuesday, February 5, 2013 at 9:21amFirst observation point A

second observation point B

top of tower P

bottom of tower Q

In triangle ABP,

angle A = 70 --- given

angle PBA = 95 --- exterior angle to 85°

angle APB = 15°

AB = 55m --- given

by the sine law

BP/sin70 = 55/sin15

BP = 55sin70/sin15

in triangle BPQ

PQ=200

angle PBQ = 85

BP = 55sin70/sin15

sinQ/BP= sin85/200

sinQ = BPsin85/200

= (55sin70/sin15)(sin85)/200 = .9946..

angle Q = 84.0658°

or appr 84°

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