I think we can use the identity sin x/2 = sqrt [(1 - cos x) /2]
cos x - sqrt3 sqrt ( 1 - cos x) /sqrt2 = 1
cos x - sqrt(3/2) sqrt(1 - cos x) = 1
sqrt(3/2)(sqrt(1 - cos x) = cos x - 1 Squaring both sides:-
1.5 ( 1 - cos x) = cos^2 x - 2 cos x + 1
cos^2 x - 0.5 cos x - 0.5 = 0
cos x = 1 , -0.5
giving x = 0 , 2pi, 2pi/3, 4pi/3 ( for 0 =< x <= 2pi)
because of thw square roots some of these solutions may be extraneous so we should plug these into the original equations to see if they fit.
The last 2 results dont fit so the answer is x = 0 , 2pi Answer
Answer:
5.07
Step-by-step explanation:
Use Pythagorean Theorem of a^2 + b^2 = c^2 where a and b are the legs of the triangle set up by the house height and the ground, and c is the hypotenuse or how long the ladder is. C is going to be our unknown.
Just plugging in you get 25^2 + 35^2 = c^2. Simplify to 625 + 1225 = c^2. Simplify again to 1850 = c^2. The square root both sides to isolate the variable c. C = sqrt(1850) or approximately 43.0116 feet if rounded to 4 decimal places.
The ladder is approximately 43.0116 feet long.
