Functions and Trigonometric Equations 52. Viewing Angle of an observer While vis
ID: 3400721 • Letter: F
Question
Functions and Trigonometric Equations 52. Viewing Angle of an observer While visiting a museum, Marsha Langlois views a painting that is 3 ft 3 ft high and hangs 6 ft above the ground See the figure. Assume her eyes are 5 ft above the ground, and let x be 6 ft 5 ft. the distance from the spot where she is standing to the wall displaying the painting. (a) Show that 0, the viewing angle subtended by the painting, is given by tan (b) Find the value of x to the nearest hundredth for each value of 0 (ii) (i) 0 (c) Find the value of to the nearest hundredth for each value of x. (ii) x (i) x Ihn Intion hflow tis time (in seconds) and y is the angleExplanation / Answer
1) from traingle 1 with base x ft and height 4 ft with angle a
tana = 4/x ; a = tan^-1(4/x)
from traingle 2 with base x ft and height 1 ft with angle b
tanb = 1/x ; b = tan^-1(1/x)
theta = a - b = tan^-1(4/x) - tan^-1(1/x)
2) i) theta = pi/6
pi/6 =tan^-1(4/x) - tan^-1(1/x)
pi/6 = tan^-1[ ( 4/x - 1/x)]/[ 1 +4/x^2]
1/sqrt3 = 3/(x +4/x)
x +4/x = 3sqrt3
x^2 - x(3sqrt3 )+4 =0
solve for x : we get x = 4.25 ft , 0.939 ft
ii) Similarly
theta = pi/8
pi/8 =tan^-1(4/x) - tan^-1(1/x)
pi/8 = tan^-1[ ( 4/x - 1/x)]/[ 1 +4/x^2]
tan(pi/8) = 3/(x +4/x)
0.414(x +4/x) = 3
0.414x^2 + -3x + 1.656 =0
solve for x : x = 6.64 ft ; x = 0.60 ft
c) i) theta = tan^-1(4/x) - tan^-1(1/x)
x =4 ; theta = tan^-1(4/4) - tan^-1(1/4)
= 45 - 14.04
= 30.96 deg
ii) theta = tan^-1(4/x) - tan^-1(1/x)
plug x = 3 ; theta = tan^-1(4/3) - tan^-1(1/3)
= 53.13 deg - 18.26 deg
= 34.87 deg
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