Question

Given 17sinalpha-15=0 and 90°<alpha<270° with the aid of the sketch using a calculator determine
a) tanalpha

Answer 1

in QII,QIII, if sinx = 15/17, then
tanx = -8/17 in QII
tanx = 817 in QIII

Answer 2

my bad. I did indeed give cosx, not tanx
if sinx = 15/17, then
tanx = -15/8 in QII
tanx = 15/8 in QIII

Answer 3

To find the value of tan(alpha), we first need to solve for alpha in the equation 17sin(alpha) - 15 = 0.

Adding 15 to both sides, we get:
17sin(alpha) = 15

Dividing by 17:
sin(alpha) = 15/17

To find the value of alpha, we use the arcsin function on a calculator:
alpha = arcsin(15/17) ≈ 56.03 degrees

Now that we have found the value of alpha, we can find tan(alpha):
tan(alpha) = tan(56.03) ≈ 1.496

So, tan(alpha) ≈ 1.496.

Answer 4

Apologies for the mistake in my previous response. Let's correct that:

Given sin(x) = 15/17, we are looking for the value of tan(x) in Quadrant II and Quadrant III.

In Quadrant II, we have sin(x) = 15/17 and tan(x) is negative. Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can find cos(x):

cos(x) = ±√(1 - sin^2(x))
cos(x) = ±√(1 - (15/17)^2)
cos(x) = ±√(1 - 225/289)
cos(x) = ±√(289/289 - 225/289)
cos(x) = ±√(64/289)
cos(x) = ±8/17

Since tan(x) = sin(x)/cos(x), in Quadrant II:
tan(x) = 15/17 / (-8/17) = -15/8 = -1.875

In Quadrant III, tan(x) will be the negative reciprocal of tan(x) in Quadrant II:
tan(x) = 1 / (-1.875) = -0.5333

Therefore, in Quadrant II, tan(x) = -8/17 and in Quadrant III, tan(x) = -0.5333. Thank you for pointing out the error.

Answer 5

No problem at all! Here is the correct calculation for tan(x) in Quadrant II and Quadrant III given sin(x) = 15/17.

In Quadrant II:
sin(x) = 15/17 (positive in Quadrant II)
cos(x) = -√(1 - sin^2(x)) = -√(1 - (15/17)^2) = -√(1 - 225/289) = -√(64/289) = -8/17
tan(x) = sin(x)/cos(x) = (15/17) / (-8/17) = -15/8

So, in Quadrant II, tan(x) = -15/8.

In Quadrant III:
sin(x) = 15/17 (positive in Quadrant II)
cos(x) = -√(1 - sin^2(x)) = -√(1 - (15/17)^2) = -√(1 - 225/289) = -√(64/289) = -8/17
tan(x) = sin(x)/cos(x) = (15/17) / (8/17) = 15/8

Therefore, in Quadrant III, tan(x) = 15/8. Thank you for correcting that and allowing me to provide the accurate solution.