A) 45
B) 30
C) 50
D) 80
Correct Answer: A
Solution :
In \[\Delta 's\] ABC and ABD, \[\angle A\] is common and \[\angle ABC=\angle ADB\] Hence \[\Delta \,ABC\] and \[\Delta \,ABD\] are equiangular \[\therefore \] \[\frac{AD}{AB}=\frac{AB}{AC}\] or \[AD=\frac{A{{B}^{2}}}{AC}=\frac{A{{C}^{2}}-B{{C}^{2}}}{AC}\] \[=AC\left[ 1-{{\left( \frac{BC}{AC} \right)}^{2}} \right]\] \[=\frac{BC}{\cos {{30}^{o}}}[1-{{(\cos {{30}^{o}})}^{2}}]\] \[=\frac{100}{(0.8)}[1-{{(0.8)}^{2}}]\] \[=\frac{100\times 10}{8}\left[ 1-\frac{64}{100} \right]\] \[=\frac{100\times 10}{8}\times \frac{36}{100}=45\]You need to login to perform this action.
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