Please help - Applied trigonometry (1 Viewer)

Pace_T

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I don't know what I'm doing wrong, but I just can't work out the first three questions from fitzpatrick :confused: Can anyone please help me?

1. P,A,B,C are four points in a plane such that angles BPA and CPA are obtuse and on opposite sides of PA. PA = 8cm, BP=10cm, PC=12cm, AB=14cm and AC=18cm. Calculate the length of BC and the area of the triangle ABC.

2. P,A,B,C are four points in order on a straight road that runs up a hillside and makes a constant angle of 10degrees with the hoizontal. A flagpole whose height is h meters stands at P. From A and B, the top of the flagpole has elevations of 30 degrees and 5 degrees respectively above the horizontal. If AB is 100m long, what is the height of the flagpole?
If BC is also 100m long, what is the elevation of the top of the flagpole from C?

3. From a point, P, a man observes that the angle of elevation of the top of a cliff, A, is 40 degrees. After walking 100 m towards A along a straight rode, inclined upwards at an angle of 15 degrees to the hoizontal, the nalge of elevation of A is observed to be 50 degrees. Find the vertical height of A above P.

All your help is greatly appreciated!!!
 

FinalFantasy

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1. P,A,B,C are four points in a plane such that angles BPA and CPA are obtuse and on opposite sides of PA. PA = 8cm, BP=10cm, PC=12cm, AB=14cm and AC=18cm. Calculate the length of BC and the area of the triangle ABC.

(BP)²=14²+8²-2(14)(8)cos (angle BAP)
(10)²=.....
cos(angle BAP)=5\7
angle BAP=cos^-1 (5\7)
(PC)²=8²+18²-2(8)(18)cos(angle CAP)
(12)²=....
cos(angle CAP)=61\72
angle CAP=cos^-1 (61\72)
(PA)²=12²+18²-2(12)(18)cos(angle PCA)
(8)²=...
cos(angle PCA)=101\108
angle PCA=cos^-1 (101\108)
(PA)²=10²+14²-2(10)(14)cos(angle PBA)
cos(angle PBA)=29\35
angle PBA=cos^-1 (29\35)
using angle sum of triangle for triangle BPA, angle BPA=102 degrees approx
similarly angle CPA=127 degrees
.: angle BPC=131

now (BC)²=10²+12²-2(10)(12)cos 131
BC=20 approx

area of triangle just use 1\2*18*14*sin angle BAC
 

Pace_T

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FinalFantasy said:
using angle sum of triangle for triangle BPA, angle BPA=102 degrees approx
similarly angle CPA=127 degrees
.: angle BPC=131
hey thanks for ur help
but i dont get this part
i can see that 131 + 127 + 102 = 360 but i dont know how you done this?
in my diagram angleBPC =angle CPA- angle BPA
thanks
 

FinalFantasy

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u gota draw it such that angles BPA, CPA and BPC are angles at a point
 

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