A is known to be 6,500 feet above sea level; AB = 600 feet. The angle at A looking up at P is 20°. The angle at B looking up at P is 35°. How far above sea level is the peak P? Find the height of the mountain peak to the nearest foot. Height above sea level = ft. (Hint: Draw a perpendicular from B to . Label the various angles. Compute BJ, then BP, then PQ, rounding to the nearest tenth in each step; and finally find 6,500 + PQ.)
Is there a diagram? because i dunno where B is located?
Whats that?
Is there like a picture? Because without it, i dont know whr B is. and i wouldnt be able to solve
like this
yea at last, yea the diagram is damn important in this case. gimme a while.
thanks for coming back.
Hmm im not sure if the hint is the way i will solve the question. But here is my method. notice that the 2 triangles ( APQ and BPQ) share the same height that of PQ
Yeah
realise that tan (angle) = opp/adj. In this case the opposite side is the height PQ. Now let x be length of BQ. so BQ = x and AQ = x+600 tan 20degrees = oppo/(x+600) tan 35 degrees = oppo/ x Equate the oppo to each other. to get (x+600) tan 20 = x tan 35 (x+600) 0.364 = x(0.7) 0.364 x +218.4 = 0.7 x 218.4 = 0.336x x= 650 feet now you got x, you know BQ is 650 ft remember, tan 35degrees = oppo /x tan 35 = oppo/ 650 oppo = tan 35 * 650= 455 ft PQ is 455ft. Since A is 6500 ft above sea lvl. The peak P is 455+6500 = 6955 ft above sea level.
Thanks dude alot you have no idea how long I was stuck on that question. Do you think you can help all night and finish this geometry? Because this is the last day I can do geometry.
Oh if you have more questions, no problem.i can help with them. im free.
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There is no need to praise me, Im just glad to help since im free. Fire away your questions. I will be happy as long as you understand how to get the answers.
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no way i will accept that. Its alright. really.
Are you sure?
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