Question number 18

Q18. Given an acute triangle ABC such that sin C =  4 5 ,   tan   A = 24 7  and AB = 50. The area of the triangle ABC 

(A) 600               (B) 1200                        (C) 1800                     (D) 2400

Dear student,

tanA = BDAD=247AD=7BD24 ...1sinA = 24242+72 = 2425=BDABBD = 2425AB=24×5025=48sinC = BDBC=45cosC = 35tanC=43=BDDCDC=34BD = 34×48 = 36AD=7BD24 = 724×48=14area =12AC×BD = 12AD+DC×BD = 1236+14×48  = 1200
Regards

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