Question

The path of a satellite orbiting the earth causes it to pass directly over two tracking stations A and B, which are 46 miles apart. When the satellite is on one side of the two stations, the angle of elevation at A is 88° and the angle of elevation at B is 82'. Round all answers to 3 decimal places. x How far is the satellite from station A? miles How high is the satellite above the ground? miles
plz explain steps.

Answer 1

c (satellite) 88° 82 D A B the Heme AC is the distance belween Satellite and the station A. CD is the distance between the and Satellite ground. Hene AB = L CAD 46 miles 889 and LCBA = 82 Hene AACD and ABCD are right angle triangle det, CA = h aud AD = 2 law use We tangent From A ACH - CA AD Base Han (884) perpendicular ► tam (884) h -> h. Jan (88°) xx x (°)

from ABCD law, ook we use tangent tokiem You (82°) DB >> tan(83) h G+46) 'DB = DA + AB art 46 ch Han (82) x (x +46) (14) Since h is the So'; from (0) tam (88°) xa 219062d same from both triangle, and Ciolo weng geti, lam(82*) x(x + 6) sonsori tam (884) tan (824) x+46 a Jan 88 28.636 tau 82 = 7.115 + 46 28.636 7.115 x 4025 + 46 2 4,025 - 1 46 3.025 46 a > 46 X 15.207 mi 3. 025 from i we put the value of a then we get bort h Han (98) X:15.207 28.636 x 15.2017 cloth: 435.468 mins . The Satellite is approximately 435.468 miles som above the ground.

Now we findt ACITOS From AACD, t we I use sine law. Sin (884) гер ! AC 20 of 2d 103 AC [sin 88' 0.999 1.9997 0.999 = 435.468 Ас. AC = 435.468 0.999 435.9 84 mile 435.904 miles ', The satellite is approximately away from tracking station A. is the Satellite from ث 435.904 miles far Station A is the satellite above () 435.468 miles high the ground