12.112. The distance from the center of the base of a cone to its generatrix is $d$. The angle between the generatrix and the height is $\alpha$. Find the total surface area of the cone.
## Solution.

The total surface area of the cone $S_{\text {pol }}=S_{\text {bok }}+S_{\text {osn }}=\pi R l+\pi R^{2}$ (Fig. 12.112). From $\triangle O B A\left(\angle O B A=90^{\circ} ; \angle B O A=\angle A S O=\alpha\right) \quad R=O A=\frac{d}{\cos \alpha}$. From $\triangle S O A \quad\left(\angle S O A=90^{\circ}\right) \quad S A=\frac{S O}{\cos \alpha} ;$ from $\quad \triangle S B O \quad S O=\frac{d}{\sin \alpha}$. Then $l=S A=\frac{d}{\sin \alpha \cos \alpha}$. Hence,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-798.jpg?height=562&width=466&top_left_y=63&top_left_x=159)

Fig. 12.111

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-798.jpg?height=524&width=418&top_left_y=82&top_left_x=751)

Fig. 12.112

$$
\begin{aligned}
& S_{\text {pol }}=\pi\left(\frac{d^{2}}{\sin \alpha \cos ^{2} \alpha}+\frac{d^{2}}{\cos ^{2} \alpha}\right)=\pi \frac{d^{2}}{\cos ^{2} \alpha}\left(\frac{1}{\sin \alpha}+1\right)= \\
& =\pi \frac{d^{2}(1+\sin \alpha)}{\sin \alpha\left(\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}\right)^{2}}=\frac{\pi d^{2}\left(\cos ^{2} \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}\right)}{\sin \alpha\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{2}\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)^{2}}= \\
& =\frac{\pi d^{2}}{\sin \alpha} \frac{\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)^{2}}{\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{2}\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)^{2}}=\frac{\pi d^{2}}{\sin \alpha\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{2}}= \\
& =-\frac{\pi d^{2}}{2 \sin \alpha\left(\frac{1}{\sqrt{2}}\right)^{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{2}}=\frac{\pi d^{2}}{2 \sin \alpha\left(\frac{1}{\sqrt{2}} \cos \frac{\alpha}{2}-\frac{1}{\sqrt{2}} \sin \frac{\alpha}{2}\right)^{2}}= \\
& =\frac{\pi d^{2}}{2 \sin \alpha\left(\sin \frac{\pi}{4} \cos \frac{\alpha}{2}-\cos \frac{\pi}{4} \sin \frac{\alpha}{2}\right)^{2}}=\frac{\pi d^{2}}{2 \sin \alpha \sin ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)} \\
& \text { Answer: } \frac{\pi d^{2}}{2 \sin \alpha \sin ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-799.jpg?height=450&width=440&top_left_y=127&top_left_x=170)

Fig. 12.113

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-799.jpg?height=500&width=406&top_left_y=78&top_left_x=749)

Fig. 12.114