An aircraft leaves A and flies 257km to B on a bearing of 257 degrees. it then flies to C 215 km away on a bearing of 163 degrees from B. Calculate <ABC?

1 Answer
Mar 17, 2018

#angle A=41^@32'57'',angleB86^@,angle C^@=52^@27'03''#

Explanation:

Bearing# A -B =257^@#

Bearing# B -C =163^@#

#:.163^@+180^@=# bearing#color(magenta)(C-B= 343^@#

# :.color(magenta)(343^@-257^@=86^@ =angle B#

or

Bearing#color(magenta)( B - A =257^@-180^@= 77^@#

#:.color(magenta)(163^@-77^@=86^@ =angle B#

Let coords of# B# be# color(blue)(x 100.00# #color(blue)(y 100.00 #

#:.B-A=77^@# distance #257km#

#:.cos77^@xx257=57.812#

#:.sin77^@xx257=250.413#

:.coords #color(blue)(B=x 100.00# #color(blue)( y 100.00#

#:.x100.00+57.812=x157.812#

#:.y100.00+250.413=y350.413#

:.coords#color(magenta)(A=x157.812# #color(magenta)( y 350.413#

#:.B-C=163^@# distance #215km#

:.coords #color(blue)(B=x 100.00# #color(blue)( y 100.00#

#:.cos163^@xx215=-205.606#

#:.sin163^@xx215=62.860#

#:.x100.00-205.606=x-105.606#

#:.y100.00+62.860=y162.860#

:.coords#color(magenta)(C=x-105.606# # color(magenta)(y 162.860#

:.distance #A-C sqrt((x_2-x_1)^2+(y_2-y_1)^2#

:.distance #A-C sqrt((-105.606-157.812)^2+(162.860-350.413)^2#

#:.=sqrt((-263.418)^2+(-187.553)^2)#

#:.=sqrt(69389.043+35176.128)#

#:.=sqrt(104565.1707)#

#:.color(magenta)(=323.365km#

#:.(sinA)/215=(sin86^@)/323.365#

#:.sinA=(sin86^@xx215)/323.365#

#:.sin A=0.663263713#

#:.color(magenta)(angle A=41^@32'57''#

#:.sinC^@/257=(sin86^@)/323.365#

#:.sinC^@=(sin86^@xx257)/323.365#

#:.sinC^@=0.792831509#

#:.color(magenta)(angleC=52^@27'03''#

#:.86^@+41^@32'57''+52^@27'03''=180^@#