Word Problem?

Question

Need to know how to set up the two equations for these problems so I can work them out.

1. A metal alloy is 25% copper. Another metal alloy is 50% copper. How much of each alloy should be used to make 1000 grams of a metal alloy that is 45% copper?

2. The base of a triangle is 4 inches longer than the length of one of the other sides. Find the length of each side of the triangle.

Answer 1

a is the mass of alloy with 25% copper
b is the mass of alloy with 50%copper
a+b=1000
a*25+b*50=1000*45
b=1000-a
25a+50*1000-50a=1000*45

For part 2, If the triangle is a right triangle (a special case), then the sides are:
b
b-4
sqrt(2b^2-8b+16)

If it is not a right triangle, then you have to also know one angle. If you know the angle subtended between the base and the side that is length b-4, call it a, then you can compute the height, h of the triangle as
h=(b-4)*sin(a)
and the base of the right triangle formed by the height and b-4 as the hypotenuse, call it x
x=(b-4)*cos(a)
Since the base of the triangle is known to be b,
then the base of the right triangle with the unknown side is b-x
or
b-(b-4)*cos(a)

It shares the height of the triangle as a side with the unknown side as the hypotenuse

So, using Pythagorean theorem,
S, the unknown side, can be computed as
S^2=
(b-(b-4)*cos(a))^2+
((b-4)*sin(a))^2


j

Answer 2

.25(x) + .5(1000-x) = .45(1000)

where x is the weight of the metal alloy containing 25% copper.

I'm not as good with triangles because there are too many cosines and sines to remember.

Answer 3

1 addem
2 4/2