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Q.

In triangle ABC,AD is perpendicular to BC and AD2=BD×DC  then find the value of angle BAC.


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a

90

b

80

c

70

d

60 

answer is A.

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Detailed Solution

In the given figure,
ADBC
So,
ABD= right angle triangle  
ACD= right angle triangle
ABD=ADC=90
From Pythagoras theorem in right angled triangle ADC, 
Hypotenuse 2= Perpendicular 2+ Base 2
AB2=AD2+BD2
AD2=AB2-BD2           ….(1)
From Pythagoras theorem in right angled triangle ABD,
AC2=AD2+CD2
AD2=AC2-CD2…………(2)
Add equation (1) and equation (2),
AC2-CD2+AB2-BD2=2AD2
BD2+CD2+2(BD×DC)
BD2+CD2+2(BD×DC)=AB2+AC2
(BD+DC)2=AB2+AC2
BC2=AB2+AC2                         [since, BD+DC=BC]
Hence, triangle BAC is a right triangle from right angle A.
BAC=90l
Correct option is 1.
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