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Question Number 95326 by bobhans last updated on 24/May/20

$$\sin {72}^{\mathrm{o}}=\mathrm{p}\sqrt{3}\cos {48}^{\mathrm{o}}$$$$\mathrm{find}\tan {12}^{\mathrm{o}}?$$

Commented by mr W last updated on 24/May/20

$$\sin 72°=\sin (60°+12°)=\frac{1}{2}(\sqrt{3}\cos 12°+\sin 12°)$$$$\cos 48°=\cos (60°-12°)=\frac{1}{2}(\cos 12°+\sqrt{3}\sin 12°)$$$$\sin {72}^{\mathrm{o}}=\mathrm{p}\sqrt{3}\cos {48}^{\mathrm{o}}$$$$\frac{1}{2}(\sqrt{3}\cos 12°+\sin 12°)=p\sqrt{3}\frac{1}{2}(\cos 12°+\sqrt{3}\sin 12°)$$$$\sqrt{3}+\tan 12°=p\sqrt{3}(1+\sqrt{3}\tan 12°)$$$$\sqrt{3}(1-p)=(3p-1)\tan 12°$$$$\Rightarrow \tan 12°=\frac{(1-p)\sqrt{3}}{3p-1}$$

Commented by bobhans last updated on 24/May/20

$$\mathrm{yes}...$$

Commented by bobhans last updated on 24/May/20

$$\mathrm{how}\mathrm{with}\mathrm{qn}95260$$

Commented by bobhans last updated on 24/May/20

$$\mathrm{sir}\mathrm{jagoll}\mathrm{asking}\mathrm{to}\mathrm{you}$$

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