(05.02 LC) Given triangle ABC, which equation could be used to find the measure of ∠C? right triangle ABC with AB measuring 6, AC measuring 3, and BC measuring 3 square root of 5

A. cos m∠C = 2 square root of 5 all over 5
B. sin m∠C = square root of 5 over 5
C. cos m∠C = square root of 5 over 2
D. sin m∠C = 2 square root of 5 all over 5

D. sin m∠C = 2 square root of 5 all over 5

Step-by-step explanation:

The triangle ABC is a right angled triangle. The angle ∠C is where we are asked to find.  The side AB measures 6 , AC measures 3 and the side BC measures 3√5.

The longest side of a right angle triangle is the hypotenuse, that means the side with length  3√5 is the hypotenuse.

Therefore side BC is the hypotenuse side and we are asked to find the angle C. AB is the opposite side while AC is the adjacent side.

Using

sin C = opposite/hypotenuse = 6/3√5 = 2/√5 = 2√5/5

cos C = adjacent/hypotenuse = 3/3√5 = 1/√5

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,