Question Number 222639 by Mingma last updated on 03/Jul/25
Answered by gabthemathguy25 last updated on 03/Jul/25
$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{8}\:{cm}=\mathrm{4}\:{cm} \\ $$$$\mathrm{Slant}\:\mathrm{height}\:=\:\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\sqrt{\mathrm{100}+\mathrm{16}}=\sqrt{\mathrm{116}}\approx\mathrm{10}.\mathrm{77}\:\mathrm{cm} \\ $$$$\mathrm{cos}\left(\theta\right)=\frac{\mathrm{10}.\mathrm{77}^{\mathrm{2}} +\mathrm{10}.\mathrm{77}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}\centerdot\mathrm{10}.\mathrm{77}\centerdot\mathrm{10}.\mathrm{77}} \\ $$$$\mathrm{cos}\:\left(\theta\right)=\frac{\mathrm{10}.\mathrm{77}^{\mathrm{2}} +\mathrm{10}.\mathrm{77}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}\centerdot\mathrm{116}}=\frac{\mathrm{168}}{\mathrm{232}}\approx\mathrm{0}.\mathrm{724} \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{724}\right)\approx\:\mathrm{43}.\mathrm{86}° \\ $$
Answered by mr W last updated on 03/Jul/25
Commented by mr W last updated on 03/Jul/25
$${OA}={OB}=\frac{\mathrm{8}}{\:\sqrt{\mathrm{2}}}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${AC}=\sqrt{\mathrm{10}^{\mathrm{2}} +\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{33}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{0}.\mathrm{5}×{AB}}{{AC}}=\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{33}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{33}}} \\ $$$${BD}=\mathrm{8}\:\mathrm{sin}\:\alpha=\mathrm{8}×\frac{\sqrt{\mathrm{29}}}{\:\sqrt{\mathrm{33}}}=\frac{\mathrm{8}\sqrt{\mathrm{29}}}{\:\sqrt{\mathrm{33}}} \\ $$$$\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{{OB}}{{DB}}=\frac{\mathrm{4}\sqrt{\mathrm{2}}×\sqrt{\mathrm{33}}}{\:\mathrm{8}\sqrt{\mathrm{29}}}=\sqrt{\frac{\mathrm{33}}{\mathrm{58}}} \\ $$$$\Rightarrow\theta=\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{\mathrm{33}}{\mathrm{58}}}\approx\mathrm{97}.\mathrm{928}°\:\:\:\checkmark \\ $$
Commented by Mingma last updated on 03/Jul/25
Perfect ��