Triangle abc hac three sides with these lengths: ab=9, bc=40, and ca=41. what is the value of cos c

Step-by-step explanation:

We have a (right-angled) triangle ABC with the following side lengths:

AB=9,

BC=40; and

CA=41.

We are to find the value of cos c. We know that ∅.

In this case, since we have to find the angle c so the AB will be the opposite, BC the base and AC will be the hypotenuse.

Therefore,

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.