Question

4) in the figure the lines AB and pq are
' intersecting at 0<ROB-90, and a:b-3:2

then find c value
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Answer 1

Answer:

$ToSimplify:sin2(x)−cos2(x)sin4(x)−cos4(x)</p><p></p><p>\color{green}{{{\large {\bf{Your\:\:Answer\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}=1}}}}}YourAnswer:sin2(x)−cos2(x)sin4(x)−cos4(x)=1</p><p>\color{yellow} {\Huge {\sf{Solution:}}}Solution:</p><p></p><p>\color{blue} {\large {\bf{Factor\:\sin ^4(x)-\cos ^4(x)}}}Factorsin4(x)−cos4(x)</p><p>\tt \color{blue} {\mathrm{Rewrite\:}\sin ^4(x)-\cos ^4(x)\mathrm{\:as\:}(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x))^2-(\cos ^2(x))^2}Rewritesin4(x)−cos4(x)as(sin2(x))2−(cos2(x))2=(sin2(x))2−(cos2(x))2</p><p></p><p></p><p></p><p>\color{fuchsia} {\normalsize {\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}}Applyexponentrule:abc=(ab)c</p><p>\color{fuchsia} {\normalsize \sin ^4(x)=(\sin ^2(x))^2}sin4(x)=(sin2(x))2</p><p></p><p>\color{fuchsia} {\normalsize =(\sin ^2(x))^2-\cos ^4(x)}=(sin2(x))2−cos4(x) =</p><p>\color{fuchsia} {\normalsize \mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}Applyexponentrule:abc=(ab)c</p><p>\color{fuchsia} {\normalsize \cos ^4(x)=(\cos ^2(x))^2}cos4(x)=(cos2(x))2</p><p></p><p></p><p>\color{fuchsia} {\normalsize =(\sin ^2(x))^2-(\cos ^2(x))^2}=(sin2(x))2−(cos2(x))2 =</p><p></p><p>\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}ApplyDifferenceofTwoSquaresFormula: x^2-y^2=(x+y)(x-y)</p><p></p><p></p><p>(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))(sin2(x))2−(cos2(x))2=(sin2(x)+cos2(x))(sin2(x)−cos2(x))</p><p></p><p></p><p></p><p>=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))(sin2(x)+cos2(x))(sin2(x)−cos2(x)) =</p><p></p><p></p><p>\color{blue} {\large {\bf{Factor\:\sin ^2(x)-\cos ^2(x)}}}Factorsin2(x)−cos2(x)</p><p>\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}ApplyDifferenceofTwoSquaresFormula: x^2-y^2=(x+y)(x-y)</p><p></p><p></p><p>\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)-\cos (x))sin2(x)−cos2(x)=(sin(x)+cos(x))(sin(x)−cos(x))</p><p></p><p></p><p></p><p></p><p>(x)=(sin(x)+cos(x))(sin(x)−cos(x))</p><p>=(\sin (x)+\cos (x))(\sin (x)-\cos (x))=(sin(x)+cos(x))(sin(x)−cos(x))</p><p></p><p>\large=(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))\ \textless \ br /\ \textgreater \ (x))(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ \large =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{\sin ^2(x)-\cos ^2(x)}=(sin2(x)+cos2(x))(sin(x)+cos(x))(sin(x)−cos(x)) \textless br/ \textgreater (x))(sin(x)+cos(x))(sin(x)−cos(x)) \textless br/ \textgreater =sin2(x)−cos2(x)(sin2(x)+cos2(x))(sin(x)+cos(x))(sin(x)−cos(x)) =</p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p>\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}ApplyDifferenceofTwoSquaresFormula: x^2-y^2=(x+y)(x-y)</p><p>\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)sin2(x)−cos2(x)=(sin(x)+cos(x))(sin(x)</p><p>(x)=(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}(x)=(sin(x)+cos(x))(sin(x)−cos(x)) \textless br/ \textgreater =(sin(x)+cos(x))(sin(x)−cos(x))(sin2(x)+cos2(x))(sin(x)+cos(x))(sin(x)−cos(x)) =</p><p></p><p></p><p></p><p>\mathrm{Cancel\:}\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}:\quad \sin ^2(x)+\cos ^2(x)Cancel \ \textless \ br /\ \textgreater \ (sin(x)+cos(x))(sin(x)−cos(x))Cancel(sin(x)+cos(x))(sin(x)−cos(x))(sin2(x)+cos2(x))(sin(x)+cos(x))(sin(x)−cos(x)):sin2(x)+cos2(x)Cancel \textless br/ \textgreater (sin(x)+cos(x))(sin(x)−cos(x))</p><p></p><p>\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)+\cos(x)Cancelthecommonfactor:sin(x)+cos(x)Cancelthecommonfactor:sin(x)+cos(x)Cancelthecommonfactor:sin(x)+cos(x)</p><p></p><p>=\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)-\cos (x))}{\sin (x)-\cos (x)}sin(x)−cos(x)(sin2(x)+cos2(x))(sin(x)−cos(x)) =</p><p></p><p></p><p>\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)-\cosCancelthecommonfactor:sin(x)−cos</p><p>\mathrm{Use\:the\:following\:identity}:\quad \cos ^2(x)+\sinUsethefollowingidentity:cos2(x)+sin</p><p>\huge \boxed{\color{red} {\ \huge =1}} =1</p><p></p><p> KAISA LAGA$

Answer 2

Answer:

Thanks for free points ☺.