Question-39303

Question Number 39303 by behi83417@gmail.com last updated on 05/Jul/18

Commented by behi83417@gmail.com last updated on 05/Jul/18

e q u a l i t e r a l t r i a n g l e w i t h s i d e s :

B C = 1, r o t a t e s a b o u t : B, a t a n g l e : 15^{∙}

1) f i n d c o m m o n a r e a b e t w e e n

r e s t a n d r o t a t e d c a s e .

2) y e s o r n o!

c o m m o n r e g i o n i s a c y c l i c s h a p e .

Answered by MrW3 last updated on 05/Jul/18

Commented by MrW3 last updated on 05/Jul/18

l e t a = s i d e l e n g t h = 1

l e t θ = r o t a t i o n a n g l e = 15 °

B D = \frac{\sqrt{3} a}{2 \cos \frac{θ}{2}}

α = 30 ° - \frac{θ}{2}

β = 60 ° + θ

\Rightarrow α + β = 90 ° + \frac{θ}{2}

\frac{E B}{\sin (180 ° - α - β)} = \frac{B D}{\sin β}

\Rightarrow E B = \frac{\sqrt{3} a \sin (α + β)}{2 \cos \frac{θ}{2} \sin β} = \frac{\sqrt{3} a}{2 \sin (60 ° + θ)}

A_{Δ D B E} = \frac{1}{2} \times D B \times E B \times \sin α

A_{c o m m o n} = 2 A_{Δ D B E} = D B \times E B \times \sin α

= \frac{\sqrt{3} a}{2 \cos \frac{θ}{2}} \times \frac{\sqrt{3} a}{2 \sin (60 ° + θ)} \times \sin (30 ° - \frac{θ}{2})

= \frac{3 a^{2} \sin (30 ° - \frac{θ}{2})}{4 \cos \frac{θ}{2} \sin (60 ° + θ)}

= A_{0} \frac{\sqrt{3} \sin (30 ° - \frac{θ}{2})}{\cos \frac{θ}{2} \sin (60 ° + θ)}

t h e c o m m o n a r e a i s n o t c y c l i c s i n c e

2 β = 120 ° + 2 θ i s n o t a l w a y s e q u a l t o 180 ° .

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