Question

An angle measuring (468n)° is in standard position. For which value of n will the terminal side fall on the x-axis?

Answer 1

Given: Measure of angle = (468 n)°

To Find: The value of n for which the terminal side fall on the x-axis

Solution:

Angle =(468 n)°

for , n=1

Angle = 468°= 360°+108°→108°> 90°(Second quadrant)

for, n=2

Angle = 468°×2=936°=720°+216°→216°> 180°(Third quadrant)

For, n=3

Angle = 468°× 3 = 1404°=1080°+324°→324°>270°(Fourth quadrant)

For, n=4

Angle = 468°× 4 = 1872°=1800°+72°→72°(First quadrant)

For, n=5

Angle = 468°× 5 = 2340°=180°×13→Falls on X axis.

Alternative Method:

(468 n)°=90° k

n=
`(90k)/(468)=(5 k)/(26)`, so if k is a multiple of 26 , we get different values of n.

Smallest value of n for which terminal side fall on the x-axis is when (k=26, n=5), (k= 52, n=10),......

Answer 2

he answer is n=5.

I got the answer through solving all ns and graphing them.

If n = 4 -> 468 (4) = 1848 degrees. It falls on the first quadrant, thus it does not fall along the negative portion of the x-axis.

If n=5 -> 468 (5) = 2310 degrees. It falls on the second quadrant, thus it falls along the negative portion of the x-axis.

If n=6 -> 468 (6) = 2808 degrees. It falls on the fourth quadrant, thus it does not fall along the negative portion of the x-axis.

If n=7 -> 468 (7) = 3276 degrees. It falls on the first quadrant, thus it does not fall along the negative portion of the x-axis.