the playground at a daycare center has a triangular-shaped sandbox. Tow of the sides measure 20 feet and 14.5 feet and form and included angle of 45 degree. Find the length and the third side of the sandbox to the nearest tenth of a foot.
Use the cosine law
i know...
a^2=b^2+c^2-2bc cos A
but i didn't get the correct answer...which is 14
is it like a^2=20^2+14.5^2-2(20)(14.5) cos135
2x2(x+6)+3x(x-5)
A^2 = 20^2+14.5^2 - 2(20)(14.5)cos45
A^ = 610.25 - 580cos(45)
A^2 = 200.14
Square root of A = square root of 200.14
A = 14.1
To find the length of the third side of the sandbox, we can use the Law of Cosines, which states:
c^2 = a^2 + b^2 - 2ab*cos(C)
In this case, we have two sides (a = 20 ft and b = 14.5 ft) and the included angle (C = 45 degrees). We need to find the length of the third side (c).
Substituting the given values into the formula, we have:
c^2 = 20^2 + 14.5^2 - 2 * 20 * 14.5 * cos(45)
Now, let's calculate this expression step by step:
c^2 = 400 + 210.25 - 2 * 20 * 14.5 * (0.7071)
c^2 = 400 + 210.25 - 580.86
c^2 = 29.39
To find c, we take the square root of both sides:
c = √29.39
c ≈ 5.4 ft
So, the length of the third side of the sandbox is approximately 5.4 feet.