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Question Number 133069 by Dwaipayan Shikari last updated on 18/Feb/21

Commented by Dwaipayan Shikari last updated on 18/Feb/21

$O n e c a n s t r e t c h a n d r e f o r m s i d e s A B a n d A C, b y$ $c o n s i d e r i n g l e n g t h B C a s c o n s t a n t . t h e a n g l e$ $ϕ c a n c h a n g e . F i n d t h e r e l a t i o n s h i p o f c h a n g e s b e t w e e n S i d e$ $A B, A C a n d a n g l e ϕ$ $(A s s u m i n g t h a t t h e t r i a n g l e w i l l n o t b r e a k, a n d l e n g t h s a r e$ $i n t e r e l a t e d)$
Commented by mr W last updated on 18/Feb/21

$B C = a = c o n s t a n t$ $A C = b = v a r i a b l e$ $A B = c = v a r i a b l e$ $∠ A = ϕ = v a r i a b l e$ $b^{2} + c^{2} - 2 b c \cos ϕ = a^{2}$ $o r$ ${(b - c)}^{2} + 2 b c (1 - \cos ϕ) = a^{2}$
Commented by Dwaipayan Shikari last updated on 18/Feb/21

$b^{2} + c^{2} - 2 b c c o s ϕ = a^{2}$ $\Rightarrow 2 b \dot{b} + 2 c \dot{c} - 2 \dot{b c c o s} ϕ - 2 c \dot{b c o s} ϕ + 2 b c s i n ϕ \dot{ϕ} = 0$ $\Rightarrow b \dot{b} + c \dot{c} + b c s i n ϕ \dot{ϕ} = c o s ϕ (b \dot{c} + c \dot{b})$ $C a n w e d o f u r t h e r l i k e t h i s ?$
Commented by mr W last updated on 18/Feb/21

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