In the adjoining figure, AB = 4 m and ED = 3 m.
If sin α = 3/5 and cos β = 12/13, find the length of BD.
Answer :
sin α = AB/AC = 3/5
AB = 3 and AC = 5
AC2 = AB2 + BC2
52 = 32 + BC2
25 = 9 + BC2
⇒ BC2 = 25 – 9 = 16
BC2 = 42
⇒ BC = 4
tan α = AB/BC = 4/5
cos β = CD/CE = 12/13
CD = 12 and CE = 13
CE2 = CD2 + ED2
132 = 122 + ED2
ED2 = 132 – 122
⇒ ED2 = 169 – 144 = 25
ED2 = (5)2
⇒ ED = 5
⇒ tan β = ED/CD = 5/12
tan α = AB/BC = 4/BC
¾ = 4/BC
⇒ BC = (4×4)/3 = 16/3 m
tan β = ED/CD = 3/CD
⇒ 5/12 = 3/CD
CD = (12 × 3)/5 = 36/5 m
Here,
BD = BC + CD
= 16/3 + 36/5
Taking LCM
= (80 + 108)/15
= 188/15 m
= 12 8/15 m
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