Construct triangle KBC in which B=120* BC=3cm and kB+KC=4.7 cm

Dear Student,

Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Step}1=\mathrm{Draw} \mathrm{a} \mathrm{line} \mathrm{BC} \mathrm{of} 3 \mathrm{cm} \mathrm{length} \\ \mathrm{Step}2=\mathrm{Draw} \mathrm{a} 120° \mathrm{line} \mathrm{BX} \\ \mathrm{Step}3=\mathrm{Mark} \mathrm{an} \mathrm{arc} \mathrm{of} \mathrm{length} \left(\mathrm{KB}+\mathrm{KC}=4.7\mathrm{cm}\right) \\ \mathrm{Step}4=\mathrm{Join} \mathrm{that} \mathrm{arc} \mathrm{to} \mathrm{C} \mathrm{and} \mathrm{bisect} \mathrm{which} \mathrm{cuts} \mathrm{BX} \mathrm{line} \mathrm{at} \mathrm{K} \\ \mathrm{Step}5=\mathrm{Join} \mathrm{K} \mathrm{to} \mathrm{C} \\ \mathrm{KBC} \mathrm{is} \mathrm{a} \mathrm{triangle} \end{gathered}$$

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