Trigonometric Equations Question 261
Question: 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]
Options:
A) $ 15^{o} $
B) $ 30^{o} $
C) $ 45^{o} $
D) None of these
Show Answer
Answer:
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{(length,of,hypotenuse)}{ \begin{aligned} & \text{(length},of,perpendicular,drawn,from, \\ & \text{ opposite vertex},to,\text{hypotenuse)} \\ \end{aligned}}=\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} $ .
BETA
