Question

4:40 Done Edit 1:33 PM Thu Jul 30 Done< myopenmath.com Question 5. Version 1*/1. Score: 0/5 Given the triangle below, find the angle A and length of sidex Note: picture is NOT drawn to scale, but you can assume an angle th that appears obtuse is obtuse. 23 17 38 4 A= 123.59663286765 degre 8.7174231406006 X Score: N/A/2.5 N/A/2.5 Question 6. Version 17/1. Score: 0/5 A pilot is flying over a straight highway. He determines the angle 6.5 km apart, to be 34 and 52®, as shown in the figure. 34° 52° 6.5 km More NOTE: The picture is NOT drawn to scale. de dictance Ethan

not sure how to get those answers

Answer 1

see that LA is From the figure we Angle sine's an obtuse From Law Sina b SinB sinc where and are the A,B,C sides of a are the a, b, c triangle angles opposite to the sides respectively a = 23, b= 17, B = 380 above Here from 23 SinA 17 23 17 an obtuse an Sin38 Sina sin38 sin (29 sinas") 56.40336 713° angle 1800-56.403367 13° = 123.59663290 Again, to find ca 180°- ( 123.5966329 +380) 18.403367 13 Now since A is so that A third angle, c SinB Sinc 2 sin(18.40336713) Sin380 8.717 4 23 1406006 .e.) x- 8.7174 23 140 600 6

e 352 A From the given Figure see that 34° 180-(34+ 52 ) 180°-P6' 94. 6.5km B Now By the sine's Law 6.5 AC [ Highway and the Sin a 4 sin B path are parallel 6.5 AC So that CA and LB sina 4 sin 52° are alternate interior le of 34° and AC 7 5.13 520 respectively So the distance of the plane from point O is 5.13 km. angle of