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How do you calculate the hypotenuse of a right triangle?
The old pythagorean theorem says, "the square of the hypotenuse is equal to the sum of the square of the other two sides." What that means is, get an accurate measurement of the two sides that touch the right angle, square them and add them together then take the square root of that number and you'll have the length of the hypotenuse. In mathematical terms it's: A2+ B2= C2. The hypotenuse of a right angled triangle is the longest side, or the one opposite the right angle. It can be calculated using a variety of methods - Pythagoras theorem : A2 + B2 = C2, where A and B are the sides adjacent to the right angle, and C is the hypotenuse. The square of A plus the square of B is equal to the square of C, so C can be calculated as √(A2 + B2). Trigonometric angle rules : Sin(e) x = o/h, cos(ine) x = a/h, tan(gent) x = o/a, where x is one of the other two angles, o is the side opposite this angle, a is the side adjacent to it, and h is the hypotenuse. Using these angle rules, you can calculate the length of any of the sides given the length of one of the other sides plus an angle. There are others, but they start to become more difficult to explain without other knowledge in triangle geometry! a2+b2=c2 /l / l / l C / l A / l / l / l /____ l B Example: /l / l / l C / l A=4 / l / l / l /___ l B=3 a2+b2=c2 42+32=c2 16+9=c2 25=c2 square root of 25= square root of c2 5=c Answer: /l / l / l C=5 / l A=4 / l / l / l /____ l B=3 Formula: a2+b2=c2 Example: A=4 B=3 C=? a2+b2=c2 42+32=c2 16+9=c2 25=c2 square root of 25= square root of c2 5=c Answer: A=4 B=3 C=5
What is the formula for the determinant of a 3 x 3 matrix?
If the matrix is { a1 b1 c1} {a2 b2 c2} {a3 b3 c3} then the determinant is a1b2c3 + b1c2a3 + c1a2b3 - (c1b2a3 + a1c2b3 + b1a2c3)
What is b square times a square?
b square times a square = b2 x a2 = (ba)2
Why the square of the sum of two numbers is different from the sum of the squares of two numbers?
call the numbers a & b (a+b)2 = a2+2ab+b2 which is greater than a2 + b2 by twice the product of the numbers. Check: say 3 and 5 32 + 52 = 9 + 25 = 34 (3 +5)2 = 64, greater by twice a x b. QED -------------------- If a and b are the numbers, then (a+b)2 = a2 + 2ab + b2, which is different from a2 + b2 (not necessarily larger). The two quantities are equal only when one (or both) of a,b is zero.
Prove that two vectors must have equal magnitude if their sum is perpendicular to their difference?
Suppose the condition stated in this problem holds for the two vectors a and b. If the sum a+b is perpendicular to the difference a-b then the dot product of these two vectors is zero: (a + b) · (a - b) = 0 Use the distributive property of the dot product to expand the left side of this equation. We get: a · a - a · b + b · a - b · b But the dot product of a vector with itself gives the magnitude squared: a · a = a2 x + a2 y + a2 z = a2 (likewise b · b = b2) and the dot product is commutative: a · b = b · a. Using these facts, we then have a2 - a · b + a · b + b2 = 0 , which gives: a2 - b2 = 0 =) a2 = b2 Since the magnitude of a vector must be a positive number, this implies a = b and so vectors a and b have the same magnitude.
