If the given point is on the terminal arm of an angle A in standard position, angle A is greater than or equal to 0 degrees and it is less than or equal to 180 degrees, find the measure of angle A, to the nearest tenth of a degree.
a) (4,3)
b) (12,5)
c) (-6,8)
d) (-24,7)
PLEASE help me!! with the steps and all i have no idea about this... thankyou so much!!
To find the measure of angle A, we need to use the given point and apply the trigonometric function known as inverse tangent (tan⁻¹) to determine the angle. Here are the steps:
Step 1: Identify the given coordinates (x, y) of the point on the terminal arm of angle A.
Step 2: Calculate the value of the ratio y/x. This ratio represents the tangent of angle A, tan(A) = y/x.
Step 3: Use the inverse tangent function (tan⁻¹) to find the measure of angle A.
Step 4: Round the result to the nearest tenth of a degree.
Let's go through each option and apply the steps:
a) For the point (4, 3), calculate tan(A) = y/x = 3/4 ≈ 0.75. Now, find the inverse tangent: A ≈ tan⁻¹(0.75) ≈ 36.9°.
b) For the point (12, 5), calculate tan(A) = y/x = 5/12 ≈ 0.4167. Using the inverse tangent: A ≈ tan⁻¹(0.4167) ≈ 22.6°.
c) For the point (-6, 8), calculate tan(A) = y/x = 8/(-6) ≈ -1.3333. Using inverse tangent: A ≈ tan⁻¹(-1.3333) ≈ -54.5°.
d) For the point (-24, 7), calculate tan(A) = y/x = 7/(-24) ≈ -0.2917. Applying inverse tangent: A ≈ tan⁻¹(-0.2917) ≈ -16.9°.
Therefore, the correct answer is:
c) Angle A measures approximately -54.5° to the nearest tenth of a degree.
Note: The measure of angle A is negative in options c) and d) because the given point lies in the third and fourth quadrants, where angles are negative in standard position.
If you are just learning this, I suggest you make a sketch of the right-angled triangle
a) and b) will be in quadrant I
c) and d) will be in quadrant II
I will do b) and c)
b) by Pythagoras, it is easy to find the hypotenuse to be 13
since cosØ = 12/13
use the cosine inverse of your calculator...
enter:
2nd
cos
(12÷13)
= to get appr 22.62° or 23°
c) hypotenuse^2 = (-6)^2 + 8^2 = 100
hypot = √100 = 10
cosØ = -6/10 = -3/5 or -.6
enter:
2nd
cos
(-.6)
=
to get 126.87° or 127°