In AB^Δ C : cos^( 2) (A )+ cos^( 2) (B )+ cos^( 2) ( C )=1 . Prove that AB^Δ C is right angled. −−−−−−−−


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Question Number 164339 by mnjuly1970 last updated on 16/Jan/22

$I n A \overset{Δ}{B C} : {c o s}^{2} (A) + {c o s}^{2} (B) + {c o s}^{2} (C) = 1 .$ $P r o v e t h a t A \overset{Δ}{B C} i s r i g h t a n g l e d .$ $- - - - - - - -$
	Answered by mindispower last updated on 17/Jan/22

	${c o s}^{2} (c) - 1 = - {s i n}^{2} (A + B)$ $= - {s i n}^{2} (A) {c o s}^{2} (B) - {s i n}^{2} (B) {c o s}^{2} (A) - \frac{s i n (2 A) s i n (2 B)}{2}$ ${c o s}^{2} (A) + {c o s}^{2} (B) - {s i n}^{2} (A) {c o s}^{2} (B) - {s i n}^{2} (B) {c o s}^{2} (A) - = 0$ $\Leftrightarrow 2 {c o s}^{2} (A) {c o s}^{2} (B) - 2 s i n (A) s i n (B) c o s (A) c o s (B) = 0$ $c o s (A) c o s (B) (c o s (A + B)) = 0$ $c o s (A) = 0 \Rightarrow A = \frac{π}{2}, O r c o s (B) = 0 \Rightarrow B = \frac{π}{2}$ $A + B = \frac{π}{2} \Rightarrow C = π - (A + B) = \frac{π}{2}$
	Commented by mnjuly1970 last updated on 18/Jan/22

	$I n A \overset{Δ}{B C} : {c o s}^{2} (A) + {c o s}^{2} (B) + {c o s}^{2} (C) = 1 .$ $P r o v e t h a t A \overset{Δ}{B C} i s r i g h t a n g l e d .$ $- - - - - - - -$ $t h a n k y o u v e r y m u c h s i r . . .$ $g r a t e f u l$
	Commented by mindispower last updated on 19/Jan/22

	$w i h e P l e a s u r h a v e a n i c e d a y$
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