Functions and Trigonometric Equations 52 Viewing Angle of an

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 angle

Solution

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