Eureka Math Algebra 2 Module 2 Lesson 6 Example Answer Key
Example 1.Suppose that point P is the point on the unit circle obtained by rotating the initial ray through 30°. Find tan.Answer: What is the length QQ of the horizontal leg of OPQ?By remembering the special triangles from Geometry, we have OQ = \.
What is the length QP of the vertical leg of OPQ?Either by the Pythagorean theorem, or by remembering the special triangles from Geometry, we have QP = \.
What are the coordinates of point P?\\)
What are cos and sin?cos = \, and sin = \.
What is tan?tan = \ With no radicals in the denominator, this is tan = \.
Eureka Math Algebra 2 Module 2 Lesson 6 Opening Exercise Answer Key
Let P be the point where the terminal ray intersects the unit circle after rotation by degrees, as shown in the diagram below.
a. Using triangle trigonometry, what are the values of x and y in terms of ?Answer:x = cos, and y = sin
b. Using triangle trigonometry, what is the value of tan in terms of x and y?Answer:tan = \
c. What is the value of tan in terms of ?Answer:tan = \}\right)}\)
Illustrative Mathematics Algebra 1 Unit 6 Answer Key
The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics. Explain each step in solving a simple equation in one variable A-REIA1.
6 1 6 Illustrative Mathematics Grade 6 Unit 1 Lesson 6 Morgan Youtube
Eureka Math Algebra 2 Module 2 Lesson 6 Problem Set Answer Key
Question 1.Label the missing side lengths, and find the value of tan in the following right triangles.a. = 30
Answer:
Question 2.Let be any real number. In the Cartesian plane, rotate the initial ray by degrees about the origin. Intersect the resulting terminal ray with the unit circle to get point P.a. Complete the table by finding the slope of the line through the origin and the point P.Answer:
b. Explain how these slopes are related to the tangent function.Answer:The slope of the line through the origin and P is equal to tan.
Question 3.Consider the following diagram of a circle of radius r centered at the origin. The line l is tangent to the circle at S, so l is perpendicular to the x-axis.a. If r = 1, then state the value of t in terms of one of the trigonometric functions.Answer:t = tan
b. If r is any positive value, then state the value of t in terms of one of the trigonometric functions.Answer:
For the given values of r and 8, find t.c. = 30, r = 2Answer:t = 2 tan = \
d. = 45, r = 2Answer:t = 2 tan = 2 1 = 2
e. = 45, r = 2Answer:t = 2 tan = 2 3 = 23
f. = 45, r = 4Answer:t = 4 tan = 4 1 = 4
g. = 30, r = 3.5Answer:t= 3.5 tan = \
h. = 0, r = 9Answer:t = 9 tan = 9 0 = 0
I. = 90, r = 5Answer:Lines OR and l are distinct parallel lines when = 90. Thus, they will never intersect, and the line segment defined by their intersection does not exist.
j. = 60, r = 3Answer:t = 3 tan = 3 3 = 3
I. = A, r = 3Answer:
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Eureka Math Algebra 2 Module 2 Lesson 6 Exercise Answer Key
Exercise 1.For each value of in the table below, use the given values of sin and cos to approximate tan to two decimal places.Answer:
a. As -90° and > -90°, what value does sin approach?Answer:
b. As -90° and > -90°, what value does cos approach?Answer:0
c. As -90° and > -90°, how would you describe the value of tan = \}\right)}\)Answer:
d. As 90° and < 90°, what value does sin approach?Answer:
e. As 90° and < 90°, what value does cos approach?Answer:0
f. As 90° and < 90°, how would you describe the behavior of tan = \}\right)}\)Answer:
g. How can we describe the range of the tangent function?Answer:The range of the tangent function is , which is the set of all real numbers.
Exercise 2.Let P be the point on the unit circle with center O that is the intersection of the terminal ray after rotation by degrees as shown in the diagram. Let Q be the foot of the perpendicular line from P to the x-axis, and let the line l be the line perpendicular to the x-axis at S. Let R be the point where the secant line OP intersects the line f. Let m be the length of \.
a. Show that m = tan.Answer:Segment RS has length m, and side \ adjacent to ROS has length 1, so we use tangent:tan = \ = mThus, m = tan.
c. Why can you use either triangle, POQ or ROS, to calculate the length m? Answer:These triangles are similar by AA similarity hence, their sides are proportional.
Answer:m = \ = \