(a) Look at the student’s answer for the length CP. Using what you know about right-angled triangles, and without performing any further calculations, explain how you know that this calculated length must be wrong. The student's incorrect attemptwe first calculate the angle at A, using the fact that all threeanales in a triangle add up to 180°. we haveA = 180- (90 +36) = 5+,so the anale at A is 54°.Using the Sine Rule in trianale ABP, we haveasinAsinPSoasinPx sinA108x sin S4°sin 12°= 9181....So the lensth of BP is 92 inches (to the nearest inch)The lensth of BP is the same as the leneth of AP, and trianaleACP is right-angled, so we havecos 36adj.92hypSo92CP == 13.1....cos 36°Therefore to the nearest inch, the projector should Be placedt inches away from the centre of the viewinG area. In this question, you are asked to comment on a student's attempt atanswering the question detailed below. Read the question and the student'sincorrect attempt first, then answer the questions below.The QuestionRonan is building an extension to his house to facilitate a social spaceand games room. One of his passions is watching sport on a big screen,so he has purchased a wide-lens home cinema projector that he canconnect to his laptop, instead of buying a large-screen TV. Having readthe manufacturer's guide extensively, and researched his options on theinternet for optimising picture quality, he decides that he would like tohave a viewing area equivalent to having a 120-inch TV mounted on thewall. This means that the width of the viewing area will be 108 inches.The projector has a wide lens that covers an angle of 72°. He needs toposition the projector centrally with respect to the viewing area, on astand that you may assume is at the correct height for optimum picturequality.Ronan sketches the situation in Figure 7, where all lengths are measuredin inches. The point P represents the projector's position on the stand.Points A and B are the furthest reaches of the viewing area (the width).The point C is the centre point of the screen's width, directly in front ofthe projector. The length of PA is equal to the length of PB. AngleAPB is 72°, so angle APC is 36°.108B36Figure 7Assuming that all aspects of picture quality are satisfactory, how far fromcentre point C should the projector be placed in order to fill the fullscreen width of 108 inches?

The given problem is solved below in detail.

(a) Look at the student’s answer for the length CP. Using what you know about right-angled triangles, and without performing any further calculations, explain how you know that this calculated length must be wrong.

(a) Look at the student’s answer for the length CP. Using what you know about right-angled triangles, and without performing any further calculations, explain how you know that this calculated length must be wrong.

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Question

100%

The student's incorrect attempt
we first calculate the angle at A, using the fact that all three
anales in a triangle add up to 180°. we have
A = 180- (90 +36) = 5+,
so the anale at A is 54°.
Using the Sine Rule in trianale ABP, we have
a
sinA
sinP
So
a
sinP
x sinA
108
x sin S4°
sin 12°
= 9181....
So the lensth of BP is 92 inches (to the nearest inch)
The lensth of BP is the same as the leneth of AP, and trianale
ACP is right-angled, so we have
cos 36
adj.
92
hyp
So
92
CP =
= 13.1....
cos 36°
Therefore to the nearest inch, the projector should Be placed
t inches away from the centre of the viewinG area.

In this question, you are asked to comment on a student's attempt at
answering the question detailed below. Read the question and the student's
incorrect attempt first, then answer the questions below.
The Question
Ronan is building an extension to his house to facilitate a social space
and games room. One of his passions is watching sport on a big screen,
so he has purchased a wide-lens home cinema projector that he can
connect to his laptop, instead of buying a large-screen TV. Having read
the manufacturer's guide extensively, and researched his options on the
internet for optimising picture quality, he decides that he would like to
have a viewing area equivalent to having a 120-inch TV mounted on the
wall. This means that the width of the viewing area will be 108 inches.
The projector has a wide lens that covers an angle of 72°. He needs to
position the projector centrally with respect to the viewing area, on a
stand that you may assume is at the correct height for optimum picture
quality.
Ronan sketches the situation in Figure 7, where all lengths are measured
in inches. The point P represents the projector's position on the stand.
Points A and B are the furthest reaches of the viewing area (the width).
The point C is the centre point of the screen's width, directly in front of
the projector. The length of PA is equal to the length of PB. Angle
APB is 72°, so angle APC is 36°.
108
B
36
Figure 7
Assuming that all aspects of picture quality are satisfactory, how far from
centre point C should the projector be placed in order to fill the full
screen width of 108 inches?

Figure in plane geometry formed by two rays or lines that share a common endpoint, called the vertex. The angle is measured in degrees using a protractor. The different types of angles are acute, obtuse, right, straight, and reflex.

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