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.
Answer:
No.
Step-by-step explanation:
A quadrilateral with 3 obtuse angles is possible. You could have 100°+100°+100°+60° quadrilateral or whatever. As long as it's inner angles add up to 360°, it is possible.
Answer:
18
Step-by-step explanation:
a = 2
b = 3
3x(2)x(3)=18
L x W = 105
L + W = 26 so L = 26 - W
substitute L = 26 - W into L x W = 105
(26 - W) x W = 105
26W - W^2 = 105
W^2 - 26W + 105 = 0
(W - 21)(W - 5) = 0
W - 21 = 0; W = 21
W - 5 = 0; W = 5
The the dimensions of the rectangle are 5 ft and 21 ft.
Double check:
The sum of the length and the width is 26 feet: 21 + 5 = 26 feet
A rectangle has an area of 105 square feet: 21 x 5 = 105 square feet
We need a slope of -1/2
2x + 4y = 1
4y = -2x + 1
y = -2/4x + 1/4
y = -1/2x + 1/4...slope of -1/2
so the answer is 2x+4y=1
