Please help
Trig: Laws of Cosines.
In Δ XYZ , side y =21, side z =4 and m < x= 80* . Find side ' x' to the nearest integer.

21

Step-by-step explanation:

law of cosines states that x^2 = y^2 + z^2 - 2yzcos<X

so x^2 = 21^2 + 4^2 - (2*21*4*cos(80)) = 457 - 29.1729 = 427.8271

x = 20.6840 which rounds to 21

x = 21 (Appropriately)

Step-by-step explanation:

Cosine rule states:

a² = b² + c² - 2ab Cos A

b² = a² + c² - 2ac Cos B

c² = a² + c² - 2ab Cos C

Where:

a, b, c are the side lengths

A,B,C are the angles.

We'll take A, B, C as X, Y, Z

and

a,b,c as x,y,z

LET'S FIRE ON!

From the question:

x = ??

y = 21

z = 4

X = 80°

Therefore:

x² = y² + z² - 2yz Cos X

x² = 21² + 4² - 2*21*4 Cos 80°

x²= 441 + 16 - 168 Cos 80°

x² = 457 - 168 * 0.174

x²= 457 - 29.232

x²= 427.768

x= √ 427.768

x = 20.683

x = 21 (Appropriately)

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

1/3 + 1/5 = 8/15.

step-by-step explanation:

hope it