Question

an underground sprinkler system is laid at an angle of 34.5 degrees to a fence. the sprinkler jets are 10 m apart and have a range of 12m.

how would i determine the length of the fence that gets wet from the sprinklers, to the nearest tenth of a metre?

Answer 1

Draw triangle ABC with A at the 34.5º vertex, B on the sprinkler at 10m from A,
and C on the fence, 12m from B. Let AC = x.

Then, by the cosine rule, cos(34.5º) = (10^2 + x^2 - 12^2)/(2*10*x)
Rearranging this to quadratic form and solving, gives x = 18.8 to 1 decimal place.

Now move point B to 20m distant from point A and point C may also move.

Now by the cosine rule, cos(34.5º) = (20^2 + x^2 - 12^2)/(2*20*x)
Solving gives x = 20.4.

That didn't move very much from the previous point C, but we need
to see what happens at the next sprinkler jet.

cos(34.5º) = (30^2 + x^2 - 12^2)/(2*30*x)
Solving this for x gives no real solutions, only imaginary ones,
so it appears that BC, in this situation, must be greater than 12m.

Thus, the answer must be x = 20.4m

Answer 2

@JADA-SHA88 thank you !