Trigonometry Problem. A wooden flagpole is embedded 3m deep....

Question

A wooden flagpole is embedded 3m deep at one corner of a concrete horizotnal slab, square in form and measuring 20ft on a side. A storm broke the flagpole at a point 1m above the slab and inclined in the direction of the diagonal. The vertical angles observed at the center and at the opposite end of this diagonal to the tip of the flagpole were 65 and 35 degrees respectively. What is the total length of the flagpole above the slab in yards/

Ans: 5.61 yards

The only difficulty I encounter in this problem is to where to put the given angles.

Answer 1

. The flagpole is in one corner of a square area 20 ft on a side so that the diagonal distance is 20√2 ft and the center of the area is at 10√2 from where the pole is located. the piece that is 1 m is 3.28 ft and the flagpole that is broken and along the diagonal is considered to be x ft in length. To find x label the problem as follows. Let the upright piece be located at pt B and the height of it at pt A, let the length of the diagonal be located at pt G and the center point at pt E Now draw line from pt E to the tip of the flagpole and call the tip pt I. Draw a line from pt G to the tip. The line EI makes a 65° angle with the line BG and the line GI makes a 35° angle with the line BG. Draw a horizontal line from pt A across EI and intersect line GI at pt H. Let the pt of intersection across IE be pt C. Draw a vertical line at pt C to line BG and label the intersection pt D and do the same at pt H to line BG and label the intersection pt F. So now x is identified as line AI. Use the following trig relationships let sin 65° = 3.28/ CE and CE= 3.62 ft. Angle IEG = angle ICH =115° and angle GIE= 30° Angle ACI= 65°. Use line of sines: sin 30°/ 10√2 = sin 35°/IE and IE= 16.22 ft. Side IC =IE- CE = 12.6ft. Tan 35°= 3.28/FG and FG = 4.68 ft. So AH =BG- FG = 23.6 ft. Next sin35°/12.6= sin 30°/ CH and CH = 10.98 ft Using this AC= AH- CH = 12.62 ft. So now the triangle containing x is the following AIC where angle ACI = 65°. Use the law of cosines to find the side AI= x. So x²= 12.6² + 12.62² - 2•12.6•12.62 cos 65°. Doing the math gives x= 13.547 ft = 4.516 yds. Add 3.28/3 =1.09 and the answer is 5.609 yds for the length of the flagpole.

Answer 2

Because of the poll's length; the 65° angle is at the 1m poll length. 1m=40cm.