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Bearing And Distance - Education - Nairaland

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Bearing And Distance by fobiflex(op): 7:52am On May 20, 2022
Please help me solve this, mathematicians in the house

Re: Bearing And Distance by fobiflex(op): 8:23am On May 20, 2022
Does this mean that no one from this forum can solve this question on bearing and distance?
Re: Bearing And Distance by obiekunie01: 8:47am On May 20, 2022
Let the starting point be B and the stopping be C as shown in the diagram below.
therefore, the angle at A will be (197-90)degree = 107degree
the angle at B is 45degree, since it is directly south west of C.
therefore the angle at C will be A+B+C=180 degree(sum angle of a triangle)
107+45+C=180
therefore, C=180-107-45=28degree
therefore, using sine rule gives;
sinA/a =sinC/c
sin107/a=sin28/4
0.9563/a=0.4695/4
0.9563/a=0.11736
therefore, a =0.9563/0.11736=8.148=8.1km(approx.)

cpd.

fobiflex:
Does this mean that no one from this forum can solve this question on bearing and distance?

Re: Bearing And Distance by fobiflex(op): 12:18pm On May 20, 2022
obiekunie01:
Let the starting point be B and the stopping be C as shown in the diagram below.
therefore, the angle at A will be (197-90)degree = 107degree
the angle at B is 45degree, since it is directly south west of C.
therefore the angle at C will be A+B+C=180 degree(sum angle of a triangle)
107+45+C=180
therefore, C=180-107-45=28degree
therefore, using sine
sinA/a =sinC/c
sin107/a=sin28/4
0.9563/a=0.4695/4
0.9563/a=0.11736
therefore, a =0.9563/0.11736=8.148=8.1km(approx.)

cpd.
you're really of great help bro, I appreciate
Re: Bearing And Distance by fobiflex(op): 11:46pm On Jul 07, 2022
obiekunie01:
Let the starting point be B and the stopping be C as shown in the diagram below.
therefore, the angle at A will be (197-90)degree = 107degree
the angle at B is 45degree, since it is directly south west of C.
therefore the angle at C will be A+B+C=180 degree(sum angle of a triangle)
107+45+C=180
therefore, C=180-107-45=28degree
therefore, using sine rule gives;

sinA/a =sinC/c
sin107/a=sin28/4
0.9563/a=0.4695/4
0.9563/a=0.11736
therefore, a =0.9563/0.11736=8.148=8.1km(approx.)

cpd.
This is the question


To start a new transport company, a business man needs at least 5buses and 10 minibuses he his not able to run more than 30 vehicle altogether. A bus takes up 3unit of parking space,a minibus takes up 1 unit of parking space and their are only 54 unit available

A. Draw a graph showing a region representing possible way of success

B. If the daily running cost of bus and a mini bus are #4500 and #2400 respectively

i. What is the objective function for the total running cost per day
ii. Find the maximum daily cost and the corresponding numbers of buses and mini buses

iii. Find the minimum daily cost and the corresponding number of buses and mini buses
Sir can you help me with this, you can send the answer to my WhatsApp directly if you don’t mind
08169243842
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