Keith is looking at a cliff. He determines that the angle of elevation to the top is 70° from where he is at. 70m away from Keith, Alan estimates the angle between the base of the cliff, himself, and Keith to be 29° while Keith estimates the angle between the base of the cliff, himself, and Alan to be 48°. What is the height, h, of the cliff to the nearest tenth of a metre? 1) 78.8m
2) 83.9m
3) 89.0m
4) 95.7m

This question completely baffled me on my homework, I still haven't done it because I don't even know how to approach it (Since it's a 3D problem)
Please help me out
Also, this homework is due in an hour so I'm starting to sweat

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  4) 95.7 m

Step-by-step explanation:

The angle at the cliff base between Alan and Keith is ...

  B = 180° -29° -48° = 103°

The law of sines will tell you the distance KB from Keith to the cliff base is

  KB/sin(A) = KA/sin(B)

  KB = KA×sin(A)/sin(B) = (70 m)sin(29°)/sin(103°) ≈ 34.82935 m

The height of the cliff is found from the tangent relation ...

  Tan = Opposite/Adjacent

  tan(70°) = height/(34.83 m)

  height = (34.83 m)tan(70°) = 95.693 m

The height of the cliff is about 95.7 m.

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Additional comment

These problems can usually be resolved into two 2-D problems. Here, we can work out the distance we need from Keith to the cliff by considering only the ground plan of the site. Then we can work out the height of the cliff only considering the vertical plane containing Keith and the base of the cliff.

The attachment shows the ground plan triangle KAB.

Step-by-step explanation:

2+x*3e-s*3

You cannot simplify any further so the answer is :

2+3ex-3s