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 P. 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?
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 P. 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?
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter11: Systems Of Equations
Section11.CT: Test
Problem 24CT
Related questions
Question
Do you think that the student’s approach to solving the problem is the
most efficient method? Give a reason for your answer.
Transcribed Image Text: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:The student's incorrect attempt
We first calculate the angle at A, using the fact that all three
angles in a triangle add up to 180°. We have
A = 180 - (90 +36) =54,
so the angle at A is 54°,
Using the Sine Rule in triangle ABP, we have
a
sinA
sin?
SO
a =
x sinA
sinP
I08
x sin 54°
%3D
sin 720
= 91.81....
So the length of BP is 92 inches (to the nearest inch).
The length of BP is the same as the length Of AP, and trianale
ACP is right-angled, so we have
adj_ CP
cos 36°
hyp
92'
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.
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