Vector Algebra Question 19
Question: ABC is an isosceles triangle right angled at A. Forces of magnitude $ 2\sqrt{2,},5 $ and 6 act along $ \overrightarrow{BC},\overrightarrow{CA} $ and $ \overrightarrow{AB} $ respectively. The magnitude of their resultant force is
[Roorkee 1999]
Options:
A) 4
B) 5
C) $ 11+2\sqrt{2} $
D) 30
Show Answer
Answer:
Correct Answer: B
Solution:
- $ R\cos \theta =6\cos 0{}^\circ +2\sqrt{2}\cos (180^{o}-B)+5\cos 270^{o} $ $ R\cos \theta =6-2\sqrt{2}\cos B $ ?..(i) $ R\sin \theta =6\sin 0{}^\circ +2\sqrt{2}\sin (180^{o}-B)+5\sin 270^{o} $ $ R\sin \theta =2\sqrt{2}\sin B-5 $ ?..(ii) From (i) and (ii), $ R^{2}=36+8{{\cos }^{2}}B-24\sqrt{2}\cos B+8{{\sin }^{2}}B $ $ +25-20\sqrt{2}\sin B $ $ =61+8({{\cos }^{2}}B+{{\sin }^{2}}B)-24\sqrt{2}\cos B-20\sqrt{2}\sin B $ $ \because $ ABC is a right angled isosceles triangle i.e., $ \angle B=\angle C=45{}^\circ $
$ \therefore $ $ R^{2} $ $ =61+8,(1)-24\sqrt{2},\cdot ,\frac{1}{\sqrt{2}}-20\sqrt{2}\cdot ,\frac{1}{\sqrt{2}} $ $ =25 $ \ $ R=5 $ .
BETA
