In a Triangle ABC , AD is perpendicular to BC ,BD is 15m,and angle B is 30 degree, find AC?

• Gven-
  bd=15,
  so,cd=15
  tan30=ad/bd
  so,ad=8.6
  ac=sqrt(ad^2+cd^2)
  =10sqrt3
  =17.3
• 11 years agoHelpfull: Yes(17) No(20)
• In a Triangle ABC,
  AD is perpendicular to BC,BD is 15m,and angle B is 30 degree
  In a right angle triangle ABD,
  BD is 15m,angle B is 30 degree,angle D is 90 degree,angle A is 60 degree.
  tan30degree=opp.side/adj.side
  1/(sqrt3)=AD/BD
  1/(sqrt3)=AD/15
  AD=15/(sqrt3)==>AD=5*(sqrt3)m
  In a right angle triangle ACD,
  Angle D is 90 degree,angle A is 30 degree.
  tan30degree=opp.side/adj.side
  1/(sqrt3)=DC/AD
  1/(sqrt3)=DC/5*(sqrt3)
  DC=5m
  In Pythagoras theorem,
  right angle triangle ACD,
  AD^2+DC^2=AC^2
  (5*(sqrt3))^2+5^2 =AC^2
  25*3+25=AC^2
  75+25=AC^2
  AC^2=100
  AC=10m
  
  
• 11 years agoHelpfull: Yes(15) No(21)
• cos30=DC/AC
  AC=15*2/sqrt3
  AC=17.32
• 10 years agoHelpfull: Yes(4) No(1)
• First draw the diagram
  B=30degre also C=30degre
  because every triange total degre 180.
  So B=30d,D=90d,A=60d ad C=30d
  BD=15m also DC=15m
  sinA=DC/AC
  AC=DC/sinA
  AC=15/sin60...bcas A=60deg
  AC=15/(sqrt3/2)
  AC=2*3*5/sqrt3
  AC=10sqrt3 and also find AD
  sinC=AD/AC
  AD=sinC*AC
  AD=sin30*10sqrt3...C=30deg
  AD=1/2*10sqrt3
  AD=5sqrt3..ad verify the ans
  PITHAKARAS THEORY
  (AD)^2+(DC)^2=(AC)^2
  (5sqrt3)^2+15^2=(10sqrt3)^2
  (25*3)+225=100*3
  75+225=300
  300=300
  
• 11 years agoHelpfull: Yes(2) No(14)
• First draw the diagram
  B=30degre also C=30degre
  because every triange total degre 180.
  So B=30d,D=90d,A=60d ad C=30d
  BD=15m also DC=15m
  sinA=DC/AC
  AC=DC/sinA
  AC=15/sin60...bcas A=60deg
  AC=15/(sqrt3/2)
  AC=2*3*5/sqrt3
  AC=10sqrt3 and also find AD
  sinC=AD/AC
  AD=sinC*AC
  AD=sin30*10sqrt3...C=30deg
  AD=1/2*10sqrt3
  AD=5sqrt3..ad verify the ans
  PITHAKARAS THEORY
  (AD)^2+(DC)^2=(AC)^2
  (5sqrt3)^2+15^2=(10sqrt3)^2
  (25*3)+225=100*3
  75+225=300
  300=300
  
• 11 years agoHelpfull: Yes(0) No(19)
• tan 30=AD/BD
  so AD=15/(root 3)
  now, sin 90=AD/AC
  =>1=[(15/(root3)/AC]
  find AC FRM DIS
• 10 years agoHelpfull: Yes(0) No(1)
• tan30=ad/bd
  1sqrt3=ad/bd
  ad=15/sqrt3
  ad=5sqrt3
  ac^2=ad^2+cd^2
  ac^2=(5sqrt3)^2+15^2
  AC= 10SQRT3
  
• 9 years agoHelpfull: Yes(0) No(1)