The Blue Angels, the Navy's elite flying squadron, have performed for more than 427 million fans since their inception in 1946.† Suppose the two Blue Angel jets at points P and A in the diagram have elevations of 1570 feet and 1000 feet, respectively. Both are flying east toward the viewing crowd at T. From T, the measure of the angle of elevation of jet P is 5°, and the measure of the angle of elevation of jet A is 2.2°. (Round your answers to two decimal places.) (a) How far from the crowd is each of the jets?Plane P ftPlane A ft(b) Find the distance d between the nose tip of jet A and the nose tip of jet P.d = ft(c) What is the ground distance from the nose tip of each jet from the crowd?Plane P Plane A The Blue Angels, the Navy's elite flying squadron, have performed for more than 427 million fans since their inception in 1946.1Suppose the two Blue Angel jets at points P and A in the diagram have elevations of 1570 feet and 1000 feet, respectively. Both are flying east toward the viewing crowd at T. From T, the measure of the angle of elevation of jet P is 5°, and the measure of the angle ofelevation of jet A is 2.2°. (Round your answers to two decimal places.)PT(a) How far from the crowd is each of the jets?Plane PftPlane Aft(b) Find the distance d between the nose tip of jet A and the nose tip of jet P.d =ft(c) What is the ground distance from the nose tip of each jet from the crowd?Plane PPlane AjanqpeagShutterstock.com

Elevation of plane P = 1570 ft,Elevation of plane A = 1000 ft,Angle of Elevation of plane P = 5o,Angle of Elevation of plane A = 2.2o,Above conditions drawn as follows, a.We know that,from trigonometric ratios,sin x = perpendicularHypotenuse,using this relation,In ∆TPXsin 5o = PXTP,substitute all known values in above equation,we get,sin 5o = 1570 ftTP,TP = 1570 ftsin 5o,TP = 1570 ft0.08715,TP ≈ 18,013.88 ft Answer Rounded up to nearest two decimal placesthis is the distance between crowd and plane P. Now,In ∆TAYsin 2.2o = AYTA,substitute all known values in above equation,we get,sin 2.2o = 1000 ftTA,TA = 1000 ftsin 2.2o,TA = 1000 ft0.03839,TA ≈ 26,048.45 ft Answer Rounded up to nearest two decimal placesthis is the distance between crowd and plane A.b.From geometry,XY = TY - TX,using trigonometric ratios,tan 2.2o = AYTY =1000TY and tan 5o = PXTX=1570TXXY = 1000tan 2.2o - 1570tan 5o,XY = 10000.038416 - 15700.0874886,XY = 8085.56 ft = OA,now,OP = PX- OX,OP = 1570 - AY,OP = 1570 -1000,OP = 570 ftNow,using Pythagorean theorem,In ∆OPA,PA2 = OP2 + OA2,Substitute all known values in above equation,we get,d2 = 5702 + 8085.562 d = 324,900 + 65,376,280.5136 ft,d = 65,701,180.5136 ft,d ≈ 8105.63 ft AnswerThis is the required distance d between the nose tip of jet A and the nose tip of jet P.c.From geometry,In ∆TPX,tan 5o = PXTX=1570TX,TX = 1570tan 5oTX =15700.0874886,TX = 17,945.20 Answer this is the required ground distance from the nose tip of Plane P from the crowd,Now,In ∆TAY,using trigonometric ratios,tan 2.2o = AYTY =1000TY ,TY = 1000tan 2.2oTY = 10000.038416 ,TY = 26,030.82 Answer,this is the required ground distance from the nose tip of Plane A from the crowd.Based on above discussion, Answer : a. Distance from the crowd to each of the jetsPlane P : 18,013.88 ftPlane A : 26,048.45 ftb. The distance d between the nose tip of jet A and the nose tip of jet Pd = 8105.63 ftc. The ground distance from the nose tip of each jet from the crowdPlane P : 17,945.20 ftPlane A :26,030.82 ft