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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Expert Answers

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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.


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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