Latest Capgemini Aptitude Question SOLUTION: In a Triangle ABC , AD is perpendicular to BC ,BD is 15m,and angle B is 30 degree, find AC?

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

Asked In Capgemini D.Jagruthi (11 years ago)

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GUDDY  (10 years ago)

Gven-
bd=15,
so,cd=15
tan30=ad/bd
so,ad=8.6
ac=sqrt(ad^2+cd^2)
=10sqrt3
=17.3

Is this solution Helpfull? Yes (17) | No (20)  

Balakrishnan M  (11 years ago)

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

Is this solution Helpfull? Yes (15) | No (21)  

Jacky singh  (10 years ago)

cos30=DC/AC
AC=15*2/sqrt3
AC=17.32

Is this solution Helpfull? Yes (4) | No (1)  

rashik fareeth  (11 years ago)

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

Is this solution Helpfull? Yes (2) | No (14)  

rashik fareeth  (11 years ago)

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

Is this solution Helpfull? Yes | No (19)  

sibani samal  (10 years ago)

tan 30=AD/BD
so AD=15/(root 3)
now, sin 90=AD/AC
=>1=[(15/(root3)/AC]
find AC FRM DIS

Is this solution Helpfull? Yes | No (1)  

Priyanka verma  (9 years ago)

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

Is this solution Helpfull? Yes | No (1)  

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