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geometry

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In triangle ABC, BC=40m , AB=50m and Angle A=53 degree. Determine the area of a triangle.

  • geometry - ,

    find angle C first of all
    sinC/50 = sin53/40
    sinC=.99829
    angle C = 86.65° or 93.35° (we have the ambiguous case, you can actually draw two triangles satisfying your information.

    then angle B = 40.35° or 33.65°

    Case 1
    AB=50
    BC=40
    angle B = 40.35
    area = (1/2)(50)(40)sin40.35
    = 647.41

    Do the second case in the same way, you should get 554.16

  • geometry - ,

    According to the law of sines,
    40/sin 53 = 50/sin C
    Therefore
    sin C = (50/40)*sin53 = 0.998
    C = 86.7 degrees
    B = 180 - 53 - 86.7 = 40.3 degrees

    With all sides and all angles, you can solve for the area. If you pick AC as the base,
    Area = (AC)*(1/2)*(BC)*sin 86.7
    = 50*(0.5)*40*.998 = 998 m^2

  • geometry - ,

    Since my numbers don't agree with Reiny's, check mine for mistakes. Reiny is right about there being two cases. I assumed all acute angles

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