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  • Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 20 maths

Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 20 maths

Answers (1)

Answer:

To have    \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}

Hint:

To solve this we use |\vec{a} \times \vec{b}| \text { and }|\vec{a}||\vec{b}| formula

Given:

Solution  : \overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B}=0

\begin{aligned} &\overrightarrow{A B} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{A B} \times 0 \\\\ &a c(\sin B-\sin A)=0 \end{aligned}

Divide by abc

\begin{aligned} &\frac{\sin B}{b}-\frac{\sin A}{a}=0 \\\\ &\frac{\sin B}{b}=\frac{\sin A}{a} \end{aligned}                                   .......(i)

\begin{aligned} &\overrightarrow{B C} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{B C} \times 0 \\\\ &\frac{\sin B}{b}=\frac{\sin C}{c} \end{aligned}.......(ii)

\begin{aligned} &\overrightarrow{C A} \times(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=\overrightarrow{B C} \times 0 \\\\ &\frac{\sin C}{c}=\frac{\sin A}{a} \end{aligned}.......(iii)

From (i),(ii) and (iii)

 

\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}

 

 

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