Question

ABC is an isosceles triangle right angled at A. Forces of magnitude $2 \sqrt { 2,5 }$ and 6 act along $\overrightarrow { B C } , \overrightarrow { C A }$ and $\overrightarrow { A B }$ respectively. The magnitude of their resultant force is

• A. 4
• B. 5
• C. $11 + 2 \sqrt { 2 }$
• D. 30

Answer 1

Answer: (B)

5

Answer 2

$R \cos \theta = 6 \cos 0 ^ { \circ } + 2 \sqrt { 2 } \cos \left( 180 ^ { \circ } - B \right) + 5 \cos 270 ^ { \circ }$

$R \cos \theta = 6 - 2 \sqrt { 2 } \cos B$ …..(i)

$R \sin \theta = 6 \sin 0 ^ { \circ } + 2 \sqrt { 2 } \sin \left( 180 ^ { \circ } - B \right) + 5 \sin 270 ^ { \circ }$

$R \sin \theta = 2 \sqrt { 2 } \sin B - 5$ …..(ii)

From (i) and (ii),

$R ^ { 2 } = 36 + 8 \cos ^ { 2 } B - 24 \sqrt { 2 } \cos B + 8 \sin ^ { 2 } B$ $+ 25 - 20 \sqrt { 2 } \sin B$

$= 61 + 8 \left( \cos ^ { 2 } B + \sin ^ { 2 } B \right) - 24 \sqrt { 2 } \cos B - 20 \sqrt { 2 } \sin B$

$\bullet \bullet$ ABC is a right angled isosceles triangle

i.e., $\angle B = \angle C = 45 ^ { \circ }$

$\therefore$ $R ^ { 2 }$ $= 61 + 8 ( 1 ) - 24 \sqrt { 2 } \cdot \frac { 1 } { \sqrt { 2 } } - 20 \sqrt { 2 } \cdot \frac { 1 } { \sqrt { 2 } }$ $= 25$

∴$R = 5$.