Question

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. .

Answer 1

D^2 = p^2 + q^2  - 2 pq cos D

d^2 = (4.24)^2 + (4.24)^2  - 2. (4.24) (4.24) cos 68

d^2 = 2 (4.24)^2 -  2. (4.24) (4.24) cos 68

d = 4.74

so the answer will be letter. C

hope this helps

Answer 2

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