if given values in a right triangle are cos A=1/3 and a=2 find the hypotenuse c?

Question

a. 3 sqrt2/2
b. 2 sqrt2/3
c. sqrt 3/2
d. 1/3
why?

Answer 1

adjactent = 2
&
cos A = adjacent / hypoteneuse = 1/3

therefore , putting the 2 into the formula

2 / h = 1/3
hypoteneuse = 6

Answer 2

Use the Pythagorean theorem Right triangles are 3/4/5. The formula a2 + b2 = c2 History of the theorem: The statement of the Theorem was discovered on a Babylonian tablet circa 1900-1600 B.C. Whether Pythagoras (c.560-c.480 B.C.) or someone else from his School was the first to discover its proof can't be claimed with any degree of credibility. Euclid's (c 300 B.C.) Elements furnish the first and, later, the standard reference in Geometry. In fact Euclid supplied two very different proofs: the Proposition I.47 (First Book, Proposition 47) and VI.31. The Theorem is reversible which means that a triangle whose sides satisfy a2 + b2 = c2 is necessarily right angled. Euclid was the first (I.48) to mention and prove this fact.


This portion of the theorem should help resolve the issue (EWD) sign(a + b - g) = sign(a2 + b2 - c2),

where sign(t) is the signum function:

sign(t) = -1, for t < 0,
sign(0) = 0,
sign(t) = 1, for t > 0.

The theorem this page is devoted to is treated as "If g = p/2, then a2 + b2 = c2." Dijkstra deservedly finds (EWD) more symmetric and more informative. Absence of transcendental quantities (p) is judged to be an additional advantage.

Answer 3

none of the a-d answers, above
if Cos A = a/c = 1/3
and a = 2
then by substitution 2/c = 1/3
and c = 2 x 3 = 6

Answer 4

that means that 1/3 = x/sqrt(x^2+4) when x = b

-> 3x = sqrt(x^2 +4)

-> 9x^2 = x^2 +4

-> 8x^2 = 4

-> x^2 = .5

-> x = 1/ sqrt(2)

so
c = sqrt(2^2 + (1/sqrt(2))^2)

-> c = sqrt(9/2)

-> c= 3/sqrt(2)

-> c = 3 sqrt(2)/2

or "a"