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In triangle PQR,p=56 , r = 17 and m angle Q=110 Find to the nearest tenth

Sagot :

Find q, using the cosine rule

[tex]\begin{gathered} q^2=r^2+p^2-2pr\cos Q \\ \end{gathered}[/tex][tex]\begin{gathered} q^2=17^2+56^2-2\times56\times17\cos 110^{\circ} \\ q^2=289+3136-1904\times-0.34202014332 \\ q^2=3425+651.206352892 \\ q^2=4076.20635289 \\ q=\sqrt[]{}4076.20635289 \\ q=63.8451748599 \\ q=63.8 \end{gathered}[/tex]

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