1. joshua has a ladder that is 12 ft long. he will lean the ladder against a vertical wall. for safety reasons, he wants the angle the ladder makes with the ground to be no greater than 75°. is it possible for joshua to lean the ladder against the wall so that the top of the ladder is at least 11.8 ft above the ground?

show your work and draw a diagram to support your answer.

Let h=height where top of ladder touches wall and x be distance from wall to bottom of ladder...

tana=h/x

a=arctan(h/x)  and a≤75

arctan(h/x)≤75  now taking tan of both sides :P

h/x≤tan75 now we have an ugly x value that we need to get rid of:

Using the pythagorean theorem we know:

144=x^2+h^2, x^2=144-h^2, x=√(144-h^2)  now we can use this in our inequality for x

h/√(144-h^2)≤tan75

h^2/(144-h^2)≤(tan75)^2

h^2≤144(tan75)^2-h^2(tan75)^2

h^2+h^2(tan75)^2≤144(tan75)^2

h^2(1+(tan75)^2)≤144(tan75)^2

h^2≤[144(tan75)^2]/(1+(tan75)^2)

h^2≤134.353829

h≤11.5911

So he cannot have the top of the ladder 11.8 ft above the ground and not exceed a 75° angle with the ground.

I worked it the hard way just to go through the process.  However we could have used a simple trig function to see that maximum height of the top of the ladder

sin75=h/12

h=12sin75

h≈11.59 ft.  That would be the maximum height given that we did not want to exceed 75° with the ground.

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12