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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

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

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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 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?
The student's incorrect attempt we first calculate the anale 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 S40. Using the Sine Rule in trianGle ABP, we have a sinA sinP' SO a = sinP x sinA 108 x sin 54° sin 12° = 9181.... 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 triangle ACP is right-angled, so we have adj _ CP 92 hyp cos 36° SO 92 = 13.1.... cos 360 Therefore to the nearest inch, the projector should Be placed t inches away from the centre of the viewing area.
Transcribed Image Text:The student's incorrect attempt we first calculate the anale 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 S40. Using the Sine Rule in trianGle ABP, we have a sinA sinP' SO a = sinP x sinA 108 x sin 54° sin 12° = 9181.... 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 triangle ACP is right-angled, so we have adj _ CP 92 hyp cos 36° SO 92 = 13.1.... cos 360 Therefore to the nearest inch, the projector should Be placed t inches away from the centre of the viewing area.
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