Wednesday, April 15, 2009

Solve the follow Trigs Questions

1. A Laser points to the top of a cliff at an angle of 47 degrees, and another points to the foot of the cliff. The laser gives the distance to the top of the cliff as 40 metres.

i. How tall is the cliff?
ii. How far away from the base of the cliff is the laser?


2. A line needs to be pulled from the top of a pole to transport material from one site to the top of a building, but the line cannot be elevated more than 40 degrees because of specifications. The building stands 45 metres tall, and the pole is 1 metre tall, and currently 50 metres away from the base of the building.

i. Can they mount the line to the top of the pole? If they can, why? If they can't, why not and where should the pole be moved to allow them to mount the line?

ii. If they can mount the line, what is the distance between the top of the pole and the top of the building?
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8 comments:

  1. shottaApril 16, 2009 at 7:13 PM

    solution 1

    if the lazer points to the top of the cliff then with an angle, this angle is considered to be an angle of elevation.
    therefore the height of the cliff will be the opposite.

    i) sin@ = opp/hyp
    opp = hyp*sin@
    = 40sin47
    = 29.25meters

    ii) distance from the lazer to the base of the cliff will be the adjacent side
    cos@ = adj/hyp
    adj = hyp*cos@
    = 40cos47
    = 27.27

    ReplyDelete
  2. shottaApril 16, 2009 at 7:33 PM

    solution 2
    i)
    tan@ = opp/adj
    @ = tan^-1(opp/adj)
    = 41.3 degrees
    because of specifications the line cannot be elevated to more than 40 degrees.
    with the pole 50 meters away from the building the angle to the top of the building is 41.3 degrees.
    to get the material up to the top of the building whilst complying with the specifications
    tan@ = opp/adj
    adj = 40/tan40
    = 52.43m
    this is the minimum distance the pole must be from the building to make an angle of 40 degrees

    ii)
    if they could mount the line the distance from the top of the building to the top of the line:
    sin@ = opp/hyp
    hyp = opp/sin@
    = 44/sin40
    = 68.45 meters

    ReplyDelete
  3. DarkyApril 18, 2009 at 4:21 PM

    How tall is the clff?

    This can be found using the sine rule and according to SHotta it is an angle of elevation because it is to the top of a cliff and at a angle.

    ReplyDelete
  4. starApril 22, 2009 at 11:37 AM

    question 1)

    i)
    sin@ =opposite/hypothenuse
    opp=hyp*sin@
    40sin47
    =29.25m

    ReplyDelete
  5. starApril 22, 2009 at 11:39 AM

    ii)
    cos@=adj/hyp
    adj=hyp*cos@
    =40cos47
    =27.27

    ReplyDelete
  6. starApril 22, 2009 at 11:43 AM

    Question 2)

    i)tan@=opp/adj
    @=tan inverse (opp/adj)
    41.3 degrees

    tan@= opp/adj
    adj=40/tan40
    52.43m

    ReplyDelete
  7. starApril 22, 2009 at 11:45 AM

    having trouble figuring out part ii)
    question 2

    ReplyDelete
  8. sky captainApril 25, 2009 at 7:37 AM

    solution for Q#1

    when the lazer points the cliff, with an angle it is considered to be an angle of elevation.
    therefore the height of the cliff will be the opposite side.

    a). sin@ = [opp/hyp]
    opp = [hyp*sin@]
    = [40sin47]
    = [29.25 m]

    b). The distance from the top, to the base of the cliff, will be the adjacent side.
    cos@ = [adj/hyp]
    adj = [hyp*cos@]
    = [40cos47]
    = [27.27]

    ReplyDelete

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