Question Number 189339 by cortano12 last updated on 15/Mar/23
Answered by HeferH last updated on 15/Mar/23
Commented by HeferH last updated on 15/Mar/23
$$\:\mathrm{cos}\:\left(\mathrm{90}°+\alpha\right)=? \\ $$$$\:\mathrm{8}\centerdot\mathrm{4}\:=\:\left(\mathrm{x}+\mathrm{14}\right)\mathrm{x} \\ $$$$\:\mathrm{x}=\mathrm{2} \\ $$$$\:\mathrm{sin}\:\:\alpha\:=\:\frac{\mathrm{4}}{\mathrm{16}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$$\:\alpha\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:\approx\:\:\mathrm{14}.\mathrm{4}° \\ $$$$\:\mathrm{cos}\:\left(\mathrm{104}.\mathrm{4}°°\right)\:\approx\:−\mathrm{0}.\mathrm{24} \\ $$
Commented by cortano12 last updated on 15/Mar/23
$$\mathrm{why}\:\mathrm{sin}\:\alpha=\frac{\mathrm{4}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{2}}\:? \\ $$
Commented by HeferH last updated on 15/Mar/23
$$\:\mathrm{because}\:\mathrm{i}\:\mathrm{made}\:\mathrm{another}\:\mathrm{mistake}!\:\mathrm{i}\:\mathrm{need}\:\mathrm{to}\:\mathrm{sleep} \\ $$$$\:\mathrm{better}\: \\ $$
Commented by cortano12 last updated on 15/Mar/23
$$\mathrm{my}\:\mathrm{solution} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{4}}{\mathrm{2r}}\:=\:\frac{\mathrm{2}}{\mathrm{r}} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}\alpha=\frac{\mathrm{14}}{\mathrm{2r}}=\frac{\mathrm{7}}{\mathrm{r}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \alpha=\frac{\mathrm{7}}{\mathrm{r}} \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{8}}{\mathrm{r}^{\mathrm{2}} }\:=\frac{\mathrm{7}}{\mathrm{r}}\:;\:\mathrm{r}^{\mathrm{2}} −\mathrm{7r}−\mathrm{8}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{r}−\mathrm{8}\right)\left(\mathrm{r}+\mathrm{1}\right)=\mathrm{0}\:;\:\mathrm{r}=\mathrm{8} \\ $$$$\mathrm{BD}=\sqrt{\mathrm{16}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }\:=\sqrt{\mathrm{20}.\mathrm{12}}=\mathrm{4}\sqrt{\mathrm{15}} \\ $$$$\:\Rightarrow\mathrm{cos}\:\measuredangle\mathrm{BCD}=\frac{\mathrm{16}+\mathrm{196}−\mathrm{240}}{\mathrm{2}.\mathrm{4}.\mathrm{14}} \\ $$$$\Rightarrow\mathrm{cos}\:\measuredangle\mathrm{BCD}=−\frac{\mathrm{28}}{\mathrm{4}.\mathrm{28}}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by HeferH last updated on 15/Mar/23
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