Question-97460

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

simply apply law of cosines a=400 b=300 β=105° b^2 =a^2 +c^2 −2ac cos β

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

Answered by smridha last updated on 08/Jun/20

AC^→ =BC^→ −AB^→ ∣AC∣=(√((300)^2 +(400)^2 −2(300)(400)cos(105^° ))) =558.673m AB+BC+AC=300+400+558.673 =1258.673m

$$\boldsymbol{{A}}\overset{\rightarrow} {\boldsymbol{{C}}}=\boldsymbol{{B}}\overset{\rightarrow} {\boldsymbol{{C}}}−\boldsymbol{{A}}\overset{\rightarrow} {{B}} \\ $$$$\mid\boldsymbol{{AC}}\mid=\sqrt{\left(\mathrm{300}\right)^{\mathrm{2}} +\left(\mathrm{400}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{300}\right)\left(\mathrm{400}\right)\boldsymbol{{cos}}\left(\mathrm{105}^{°} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{558}.\mathrm{673}\boldsymbol{{m}} \\ $$$$\boldsymbol{{A}}{B}+\boldsymbol{{BC}}+\boldsymbol{{AC}}=\mathrm{300}+\mathrm{400}+\mathrm{558}.\mathrm{673} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1258}.\mathrm{673}\boldsymbol{{m}} \\ $$

Commented by 1549442205 last updated on 08/Jun/20

cos 105°=−sin15°=−(√((1−cos30°)/2)) = −(√((2−(√3))/4))=−(((√3) −1)/(2(√2)))=((−((√6)−(√2) ))/4)

$$\mathrm{cos}\:\mathrm{105}°=−\mathrm{sin15}°=−\sqrt{\frac{\mathrm{1}−\mathrm{cos30}°}{\mathrm{2}}}\:= \\ $$$$−\sqrt{\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}}}=−\frac{\sqrt{\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{−\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\:\right)}{\mathrm{4}} \\ $$

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Question-97460

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