In a laboratory experiment, an ultrasound detector located at D is used to measure the distances of two moving objects, P and Q. At a certain moment, P and Q are located as shown in the diagram. What is distance PQ?. a) 6.82. b)5.66. c)4.74. d)4.24. e)3.89. .
Triangle PDQ has:
Side "p" is opposite Angle P
Side "d" is opposite Angle D = 45º + 23º = 68º
Side "q" is opposite Angle Q
Use the Law of Cosines:
d² = p² + q² − 2 • p • q • cos(D)
d² = (4.24)² + (4.24)² − 2 • (4.24) • (4.24) • cos(68º)
d² = 2(4.24)² − 2(4.24)² • cos(68º)
d² = 2(4.24)² • [ 1 – cos(68º) ]
d = 4.74 ft
The distance PQ is 4.74 feet.
Triangle PDQ has:
If we use the Law of Cosines:
d² = p² + q² − 2 • p • q • cos(D)
d² = 2(4.24)² • [ 1 – cos(68º) ]
d = 4.74 ft
Read more about distances here:
https://brainly.com/question/17273444
#SPJ5