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Question Number 160205 by Tawa11 last updated on 25/Nov/21

	Answered by mr W last updated on 26/Nov/21

	Commented by mr W last updated on 26/Nov/21

	$o r a n g e s h a d e d a r e a$ $= \frac{8}{2} \times 4 \tan 54 ° - \frac{8}{2} \times 4 \tan 36 °$ $= 16 (\tan 54 ° - \tan 36 °)$ $a r e a o f s t a r = 5 \times 16 (\tan 54 ° - \tan 36 °)$ $= 80 (\tan 54 ° - \tan 36 °) \approx 51.987 {c m}^{2}$ $\underset{―}{e x a c t f o r m :}$ $\sin 18 ° = \frac{\sqrt{5} - 1}{4}$ $\cos 36 ° = 1 - 2 {(\frac{\sqrt{5} - 1}{4})}^{2} = \frac{\sqrt{5} + 1}{4}$ $\tan 36 ° = \sqrt{{(\frac{4}{\sqrt{5} + 1})}^{2} - 1} = \sqrt{5 - 2 \sqrt{5}}$ $\tan 54 ° = \frac{1}{\tan 36 °}$ $\tan 54 ° - \tan 36 ° = \frac{1 - \tan^{2} 36 °}{\tan 36 °}$ $= \frac{1 - 5 + 2 \sqrt{5}}{\sqrt{5 - 2 \sqrt{5}}} = \frac{2 (\sqrt{5} - 2)}{\sqrt{\sqrt{5} (\sqrt{5} - 2)}} = \frac{2 \sqrt{5 (5 - 2 \sqrt{5})}}{5}$ $a r e a o f s t a r = 80 \times \frac{2 \sqrt{5 (5 - 2 \sqrt{5})}}{5}$ $= 32 \sqrt{5 (5 - 2 \sqrt{5})}$
	Commented by Tawa11 last updated on 26/Nov/21

	$God bless you sir . I appreciate$
	Answered by amin96 last updated on 25/Nov/21

	$s h a d o w a r e a = 160 \times \frac{3 - \sqrt{5}}{\sqrt{10 - 2 \sqrt{5}}}$
	Commented by Tawa11 last updated on 26/Nov/21

	$God bless you sir . I appreciate$
	Answered by 1549442205PVT last updated on 26/Nov/21

	$w e h a v e s i n 36^{0} = c o s 54^{0} \Rightarrow 2 s i n 18^{0} c o s 18^{0} = 3 c o s . 318^{0}$ $\Rightarrow 2 s i n 18^{0} c o s 18^{0} = 4 {c o s}^{3} 18^{0} - 3 c o s 18^{0}$ $\Rightarrow 4 {c o s}^{2} 18^{0} - 3 = 2 s i n 18^{0} \Rightarrow 4 (1 - {s i n}^{2} 18^{0}) - 3 = 2 s i n 18^{0}$ $\Rightarrow 4 {s i n}^{2} 18^{0} + 2 s i n 18^{0} - 1 = 0$ $\Rightarrow s i n 18^{0} = \frac{\sqrt{5} - 1}{4} . F r o m t h a t w e h a v e$ $c o s 18^{0} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$ $c o s 36^{0} = 1 - 2 {s i n}^{2} 18^{0} = 1 - \frac{3 - \sqrt{5}}{4} = \frac{\sqrt{5} + 1}{4}$ $s i n 36^{0} = \sqrt{1 - {c o s}^{2} 36^{0}} = \frac{\sqrt{10 - 2 \sqrt{5}}}{4},$ $t a n 36^{0} = \frac{\sqrt{10 - 2 \sqrt{5}}}{\sqrt{5 + 1}}$ $T h e a l t i t u d e o f e v e r y w h i t e t r i a n g l e i s e q u a l$ $h_{0} = 4 \times t a n 36^{0} = (\sqrt{5} - 1) \sqrt{10 - 2 \sqrt{5} .} H e n c e$ $t h e a r e a o f e v e r y w h i t e t r i a n g l e i s$ $s_{0} = \frac{1}{2} {m h}_{0} = \frac{1}{2} 8 . (\sqrt{5} - 1) \sqrt{10 - 2 \sqrt{5}}) = 4 (\sqrt{5} - 1) \sqrt{10 - 2 \sqrt{5}})$ $B a s a o f i t e q u a l t o b = 8 + 2.8 s i n 18 = 4 (\sqrt{5} + 1)$ $t h e a l t i t u d e o f t h e {s t a r t}^{'} s w i n g i s$ $h_{2} = 8 . s i n 36^{0} = 2 \sqrt{10 - 2 \sqrt{5}} . T h e a r e a o f e v e r y$ $t h e a l t i t u d e o f t h e t r i a n g l e w i t h t h e l a t e r a l s i d e b a n d t h e b a s e m = 8 c m$ $e q u a l t o h_{3} = b c o s 18^{0} = (\sqrt{5} + 1) \sqrt{10 + 2 \sqrt{5}}$ $t h e a r e o f t h e t r i a n g l e w i t h t h e s i d e b a n d a l t i t u d e h_{2} b e :$ $e q u a l t o s_{2} = \frac{1}{2} {b h}_{2} = \frac{1}{2} = \frac{1}{2} 4 (\sqrt{5} + 1) \sqrt{10 - 2 \sqrt{5}} = 2 (\sqrt{5} + 1) \sqrt{10 - 2 \sqrt{5 u}}$ $t h e a r e o f t h e t r i a n g l e w i t h t h e l a t e r a l s i d e b a n d a l t i t u d e h_{3} b e :$ $s_{3} = \frac{1}{2} {m h}_{3} = \frac{1}{2} . 8 . (\sqrt{5} + 1) \sqrt{10 + 2 \sqrt{5}} = 4 (\sqrt{5} + 1 \sqrt{10 + 2 \sqrt{5}}$ $t h e o f t h e p e n t a g o n e q u a l t o$ $S = s_{3} + 2 s_{2} = 4 (\sqrt{5} + 1) \sqrt{10 + 2 \sqrt{5}} + 2.2 (\sqrt{5} + 1) \sqrt{10 - 2 \sqrt{5}}$ $= 4 (\sqrt{5} + 1) . 4 \sqrt{5} = 16 (5 + \sqrt{5})$ $t h e o f t h e s t a r i s : s = S - 5 s_{0}$ $= 16 (5 + \sqrt{5)} - 5.4 (\sqrt{5} - 1) \sqrt{10 - 2 \sqrt{5}})$ $\approx 57.654$
	Commented by Tawa11 last updated on 26/Nov/21

	$God bless you sir . I appreciate$
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