Question

$ABC$ is an isosceles triangle right angled at $A$. Forces of magnitude $2\sqrt{2}$, $5$ and $6$ act along $BC,CA$ and $AB$ respectively. The magnitude of their resultant forces is

• A. $4$
• B. $5$
• C. $11+2\surd 2$
• D. $30$

Answer 1

The correct option is B

$5$

Explanation for the correct option:

Step-1: Draw the figure with forces marked on sides:

Let $R$ is the resultant force

From the figure, we get

$R\cos \theta =6\cos 0°+2\surd 2\cos (180°–B)+5\cos 270°$

$R\cos \theta =6–2\surd 2\cos B$ — $\left(i\right)$

$R\sin \theta =6\sin 0°+2\surd 2\sin (180°–B)+5\sin 270°$

$R\sin \theta =2\surd 2\sin B–5$ — $\left(ii\right)$

Step- 2: Solutioj of equation$\left(i\right)$ and $\left(ii\right)$

Square and add $\left(i\right)$ and $\left(ii\right)$:

$\begin{array}{rcl}& \Rightarrow & R^2=36+8{\cos }^{2}B–24\surd 2\cos B+8{\sin }^{2}B+25–20\surd 2\sin B\\ & =& 61+8({\cos }^{2}B+{\sin }^{2}B)–24\surd 2\cos B–20\surd 2\sin B\end{array}$

$\because ABC$ is a right angled isosceles triangle.

$\Rightarrow$$\begin{array}{rcl}\angle B& =& \angle C\\ & =& 45°\end{array}$

$\begin{array}{rcl}& \Rightarrow & R^2=61+8\times 1–24\surd 2\times \frac{1}{\surd 2}–20\surd 2\times \frac{1}{\surd 2}\\ & =& 25\end{array}$

$\Rightarrow$ $R=5$

Hence, option ‘B’ is correct.