question_answer

If in a right angled triangle the hypotenuse is four times as long as the perpendicular drawn to it from opposite vertex, then one of its acute angle is   [MP PET 1998, 2004; UPSEAT 2002]

A) \[{{15}^{o}}\]

B) \[{{30}^{o}}\]

C) \[{{45}^{o}}\]

D) None of these

Correct Answer: A

Solution :

If x is length of perpendicular drawn to it from opposite vertex of a right angled triangle, So, length of diagonal \[AB={{y}_{1}}+{{y}_{2}}\] .....(i) From \[\Delta OCB,{{y}_{2}}=x\cot \theta \] From \[\Delta OCA\], \[{{y}_{1}}=x\tan \theta \] Put the values in equation (i), then      \[AB=x(\tan \theta +\cot \theta )\] .....(ii) \[\because \] Length of hypotenuse = 4 (length of perpendicular) \ \[x(\tan \theta +\cot \theta )=4x\Rightarrow \frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta .\cos \theta }=4\] Þ \[\sin 2\theta =\frac{1}{2}\Rightarrow 2\theta ={{30}^{o}}\]or \[\theta ={{15}^{o}}\]. Trick :\[\frac{(\text{length}\,\text{of}\,\text{hypotenuse})}{\begin{align}   & \text{(length}\,\text{of}\,\text{perpendicular}\,\text{drawn}\,\text{from}\, \\  & \text{      opposite vertex}\,\text{to}\,\text{hypotenuse)} \\ \end{align}}=\frac{2}{\sin 2\theta }\] Þ \[4=\frac{2}{\sin 2\theta }\Rightarrow \sin 2\theta =\frac{1}{2}=\sin {{30}^{o}}\Rightarrow \theta ={{15}^{o}}\].