Question

In a rectangle ABCD, M and N are the midpoint of the sides BC and CD respectively. If \[\overset\frown{ANM}=90{}^\circ \] then AB:BC is
A. \[2:1\]
B. \[\sqrt{2}:1\]
C. \[\sqrt{3}:1\]
D. \[1:1\]

Answer 1

Hint: Use trigonometric identities and find the relation between them. The relation cot and tan are important here, they are reciprocal of each other and the complement of tan is cot. Then put the values and solve if the square root is converted into the square from one said to another.

Complete step-by-step answer:
Let \[AB=DC=2x\] and \[AD=BC=2y\].
In
     \[\tan \theta =\dfrac{perpendicular}{base}\]
     

     \[\tan \theta =\dfrac{y}{x}\] … (1)
In \[\Delta ADN\] , \[\tan \left( 90-\theta \right)=\dfrac{2y}{x}\] …(2)
     \[\cot \theta =\dfrac{2y}{x}\]
     \[\tan \theta =\dfrac{x}{2y}\] \[\left\{ as\,\,\cot \theta =\dfrac{1}{\tan \theta } \right\}\]
By putting the value of from equation
     \[\dfrac{y}{x}=\dfrac{x}{2y}\].
     \[{{x}^{2}}=2{{y}^{2}}\]
     \[\dfrac{{{x}^{2}}}{{{y}^{2}}}=\dfrac{2}{1}\]
     \[{{\left( \dfrac{x}{y} \right)}^{2}}=\dfrac{2}{1}\]
     \[\left( \dfrac{x}{y} \right)=\dfrac{\sqrt{2}}{1}\].
The correct option is B.

Note: Firstly, draw the neat and clear diagram name every angle or side. Then write the given and what to find. Use Trigonometric identities and formulas to solve. Then solve the value of x and y according to the problems.