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UNIT 6 : BASIC TRIGONOMETRY WITH TRIANGLES
LESSON
2: ANGLES IN STANDARD POSITION HOMEWORK
QUESTIONS (Solutions below)
Definition:
An angle is in standard position if it has its vertex at the origin
and initial arm along the positive
x-axis. The terminal arm is found by
rotating the initial arm about the origin to a terminal position in one of the
4 quadrants. The rotation is positive
if it is in the counter – clockwise direction and negative if in the clockwise
direction.
To find the trigonometric ratios of angles between 90o
and 180o, use the following rules:
2. Evaluate each of the following using reference (related acute) angles.
Give answers to 5 decimal places.
a) cos 121o b)
sin 106o c)
cot 97o d)
sin 165.4o
3. Evaluate each of the following directly
using your calculator. Give answers to
5 decimal places.
a) sin 1190 b) tan 143.6o c) sec 124.9o d) sin 148.2o
a) sin B = 0.3256 b) cos B
= -0.87921 c) csc B = 1.5894 d) cos B = 0.3685
Solutions:
Solution:
x = -6 y = 8 r = 10
2. Evaluate each of the following using reference (related acute) angles.
Give answers to 5 decimal places.
a) cos 121o b)
sin 106o c)
cot 97o d)
sin 165.4o
Solutions: [Recall to get the
reference (related acute) angle, subtract from 180o].
a) cos 121o = - cos (180o – 121o)
= - cos 59o
** 59o is the reference
(related acute) angle
= - 0.51504 ** Using your cacluator
b) sin 106o = sin (180o – 106o)
= sin 74o ** 74o is the reference
(related acute) angle
=
0.96126 ** Using your cacluator
b) sin 165.4o = sin (180o – 165.4o)
= sin 14.6o
** 14.6o is the reference (related
acute) angle
= 0.25207 ** Using your cacluator
3. Evaluate each of the following directly
using your calculator. Give answers to
5 decimal places.
a) sin 1190 b) tan 143.6o c) cos 124.9o d) sin 148.2o
Solutions:
a) sin B = 0.3256 b) cos B
= -0.87921 c) csc B = 1.5894 d) cos B = 0.3685
Solutions:
a) Since
sin B is positive, < B could be
in either the first or second quadrant.
There are two solutions.
b) Since
cos B is negative, < B will be in the second quadrant.
c) Since
csc B is positive, < B could be in either the
first or second quadrant. There are two solutions.
d) Since
cos B is positive, < B will be in the first quadrant.
