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Question Number 97460 by mathocean1 last updated on 08/Jun/20

Commented by mathocean1 last updated on 08/Jun/20

$$AB+BC+AC=...?$$

Commented by MJS last updated on 08/Jun/20

$$\mathrm{simply}\mathrm{apply}\mathrm{law}\mathrm{of}\mathrm{cosines}$$$$a=400b=300\beta =105°$$$$b^2=a^2+c^2-2ac\cos \beta$$

Answered by smridha last updated on 08/Jun/20

$$\mathit{A}\overrightarrow{\mathit{C}}=\mathit{B}\overrightarrow{\mathit{C}}-\mathit{A}\overrightarrow{B}$$$$\mid \mathit{A}\mathit{C}\mid =\sqrt{{\left(300\right)}^2+{\left(400\right)}^2-2\left(300\right)\left(400\right)\mathit{c}\mathit{o}\mathit{s}\left({105}^{°}\right)}$$$$=558.673\mathit{m}$$$$\mathit{A}B+\mathit{B}\mathit{C}+\mathit{A}\mathit{C}=300+400+558.673$$$$=1258.673\mathit{m}$$

Commented by 1549442205 last updated on 08/Jun/20

$$\cos 105°=-\sin 15°=-\sqrt{\frac{1-\cos 30°}{2}}=$$$$-\sqrt{\frac{2-\sqrt{3}}{4}}=-\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{-(\sqrt{6}-\sqrt{2})}{4}$$

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