Is side lengths 22, 42, 60 a right triangle? Yes or no?

The cosine rule will be used to identify the angles.

let c = 60, a = 42,and b = 22

a²=b²+c²-2bc.cosA

42²=22²+60²-(2*22*60.cosA)

1764=4084 - 2640.cosA

2640.cosA = 4084 - 1764

cos A = 0.8787

A = 28.5°

b²=a²+c²-2ac.cosB

22²=42²+60²-(2*42*60.cosB)

484=5364-5040.cosB

5040.cosB=5364-484

cosB = 0.9682

B = 14.5°

c²=a²+b²-2ab.cosC

60²=42²+22²-(2*42*22.cosC)

3600=2248 - 1848.cosC

1848.cosC=2248-3600

cosC=-0.7316

C = 137°

From the angles obtained, it shows no angle in the triangle was right-angled, that is, 90°. This triangle wasn't right-angled triangle.

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6