law of cosine?

Question

Problem
1. A ladder 24 ft. long is leaning against a sloping embankment. the foot of the ladder is 11 ft. from the base of the embankment and the distance from the top of the ladder down the embankment to the ground is 16 ft. what is the angle at which the embankment is inclined to the horizontal?

Answer 1

no you don't need the cosine law you can use simple trigonometry to solve this but getting the answer is bit complicated specially writing it down in here.
if ladders inclined to the horizon by "a" and the embankment is inclined by "b" then we can get these equations
16Sinb = 24Sina
3Sinb = 4sina -----(1)

16Cosb = 24Cosa - 11
16[1 - Sin^2(b)]^1/2 = 24[1 - Sin^2(a)]^1/2 - 11
and you can replace sin^2(a) by [9sin^2(b)]/4
and you get
16[1 - Sin^2(b)]^1/2 = 24{1 - [9sin^2(b)]/4}^1/2 -11
we need to get rid of square roots to get the answer.
to do that square both sides twice and eventually you'll get
quadratic equation of Sin^2(b) then you can use
x= {-b +/- (b^2 - 4ac)^1/2}/2a
to get the value of Sin^2(b). I'm not going write the steps because you can do it your self an it would be very complicated after writing it down here.
good luck on solving!

Answer 2

If I am interpreting the situation correctly, the relevant facts are that we have a triangle of sides, 24,11 and 16.
Angle A between sides 11 and 16 is given by cosine rule:-
24² = 11² + 16² - 2 x 11 x 16.cos A
352.cos A = 121 + 256 - 576
A = 124°
Required angle = (180 - 124)° = 56° is angle at which embankment is inclined to horizontal.

Answer 3

You have two right triangles. Both have height h. By the Pythagorean Theorem we have:

h² = 16² - x² = 24² - (11 + x)²

256 - x² = 576 - 121 - 22x - x²
22x = 199
x = 199/22

h² = 16² - x² = 256 - (199/22)² = 163,505 / 484
h = √(163,505 / 484) = √163,505 / 22 ≈ 18.379887

The angle θ of the embankment is found in:

tan θ = h / x = (√163,505 / 22) / (199/22)

tan θ = √163,505 / 199

θ = arctan(√163,505 / 199) = 63.796417°

Answer 4

the 24 ft. is there to confuse you.
you have embankment with height of 11 ft. and hypotenuse of 16 ft. I think you can do the rest ;-)