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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

$$\cos (90°+\alpha )=?$$$$8\cdot 4=(\mathrm{x}+14)\mathrm{x}$$$$\mathrm{x}=2$$$$\sin \alpha =\frac{4}{16}=\frac{1}{4}$$$$\alpha ={\sin }^{-1}\left(\frac{1}{4}\right)\approx 14.4°$$$$\cos (104.4°°)\approx -0.24$$

Commented by cortano12 last updated on 15/Mar/23

$$\mathrm{why}\sin \alpha =\frac{4}{16}=\frac{1}{2}?$$

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

$$\mathrm{my}\mathrm{solution}$$$$\Rightarrow \sin \alpha =\frac{4}{2\mathrm{r}}=\frac{2}{\mathrm{r}}$$$$\Rightarrow \cos 2\alpha =\frac{14}{2\mathrm{r}}=\frac{7}{\mathrm{r}}$$$$\Rightarrow 1-2\sin {}^2\alpha =\frac{7}{\mathrm{r}}$$$$\Rightarrow 1-\frac{8}{{\mathrm{r}}^2}=\frac{7}{\mathrm{r}};{\mathrm{r}}^2-7\mathrm{r}-8=0$$$$\Rightarrow (\mathrm{r}-8)(\mathrm{r}+1)=0;\mathrm{r}=8$$$$\mathrm{BD}=\sqrt{{16}^2-4^2}=\sqrt{20.12}=4\sqrt{15}$$$$\Rightarrow \cos \measuredangle \mathrm{BCD}=\frac{16+196-240}{2.4.14}$$$$\Rightarrow \cos \measuredangle \mathrm{BCD}=-\frac{28}{4.28}=-\frac{1}{4}$$

Commented by HeferH last updated on 15/Mar/23

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