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?

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
ChapterP: Prerequisites
SectionP.9: Modeling With Equations
Problem 24E
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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?
Transcribed Image Text: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?
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