ASSIGNMENT > A1-M1_ L1 Alberta Distance Learning Centre MATH 20-2. Q&A. Score 100%.

downloaded from

Scholarfriends

https://scholarfriends.com Downloaded by: writerforacademic@gmail.com | https://scholarfriends.com/singlePaper/275872/assignment-a1-m1-l1-alberta-distancelearning-centre-math-20-2-qa-score-100 Distribution of this document is illegal.

Math 20-2

A1 - MODULE 1: LESSON 1 ASSIGNMENT 12 Marks Please show your work.

(4 marks)

1. Determine the measure of angle θ to the nearest degree.

Answer: BC adjacent: 13m angle: 62° opposite: ? equation

hypotenuse: 24. 4m opposite: 17m angle: ? equation

side opposite ¿ sin (θ) = ¿ θ side opposite hypotenuse side adjacent tan (θ) = 17 ¿θ ¿ θ¿ sin (θ) = ¿ 24. 4 side opposite = 0.696721 adjacent·tan (θ) = side adjacent ·adjacen sin-¹(0.696721) ¿θ ¿ θ¿ = 44.1645 ¿ Answer: t adjacent·tan (θ) =side opposite to opposite = adjacent · tan (θ) = 13 · tan(62) AB = 24. 4m

(4 marks)

θ

The measurement of angle nearest degree is 44°.

θ

to the

2. Nasreen is designing an obstacle course that includes two ladders leaning against a pole. In order for the ladders to fit in the obstacle course, one ladder must span 1.6 m horizontally and the other must span 5 m horizontally. She already has one 4.3-m ladder. How long must the other ladder be? The answer should be rounded to the nearest tenth.

This study source was downloaded by 100000857952895 from CourseHero.com on 01-06-2023 01:05:38 GMT -06:00

https://www.coursehero.com/file/67431592/A1-M1-L1docx/

Math 20-2

; Answer: equation: a2 + b2 = c2 1.6m2 + x2 = 4.3m2 2.56m + x2 = 18.49 x2 = 18.49 - 2.56 x2 = √ ❑ x = 3.99124 x = 4m (4 marks)

y2 = 4m2 + 5m2 = 16 + 25 = 41 = √❑ y = 6.4m Answer: The other ladder must be 6.4m long.

3. The following picture shows a puzzle with a number of triangles. There are a number of ways to determine the length of side x. Your goal is to determine the length of side x using as few calculations as possible. The final answer should be rounded to the nearest tenth.

This study source was downloaded by 100000857952895 from CourseHero.com on 01-06-2023 01:05:38 GMT -06:00

https://www.coursehero.com/file/67431592/A1-M1-L1docx/

Math 20-2 Answer: adjacent: 3.6cm angle: 42° (w)opposite: ? equation:

side opposite tan (θ) = side adjacent ¿θ ¿ θ¿ ¿ side opposite to θ = tan (θ) · side adjacent to θ = tan(42)·3.6 = 3.24145 w = 3.2

adjacent: 3.2cm cos (θ) : 38° (x)hypotenuse: ? equation:

side adjacent ¿ hypotenuse side adjacent hypotenuse = ¿ θ ¿ cos (θ) 3.2 = cos(38) cos (θ) = ¿ θ

= 4.06086 x = 4.1cm Answer: The length of side x rounded to the nearest tenth is 4.1cm.

Submit your completed assignment to Dropbox A1 by the deadline.

This study source was downloaded by 100000857952895 from CourseHero.com on 01-06-2023 01:05:38 GMT -06:00

https://www.coursehero.com/file/67431592/A1-M1-L1docx/ Powered by TCPDF (www.tcpdf.org)