In the diagram,ABC represents a horizontal triangular field.A path runs along the edge AB of the field.BC=52m,AC=71m, and angle ACB =90

Find the greatest angle of elevation of the top of the tree when viewed from any point on the path AB.Thanks

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The greatest angle of elevation will occur at the point on AB that is closest to C. That point will be B.

The angle of elevation is found by using the inverse tangent of the ratio of PC to BC.

`tan /_PBC = 23/52 ==> m/_PBC=tan^(-1)(23/52)~~23.86^@`

**The greatest angle of elevation is approximately 23.9 degrees.**

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Correct answer:

The greatest angle of elevation will occur from the point on AB closest to C.

The shortest distance from a point on AB to C is approximately 42m.

(http://www.enotes.com/homework-help/diagram-abc-represents-horizontal-triangular-453131)

The angle of elevation can be found from the point on AB, Q, that is the closest to C by `tan/_PQC=23/42==> m/_PQC=tan^(-1)(23/42)~~28.7^@`

Thus the greatest angle of elevation is approximately 28.7 degrees.

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