If xsin⁡θ$$=ysin$$⁡θ$$+2$$$$π$$$$3=zsin$$⁡θ$$+4$$$$π$$$$3$$ then

Question

If xsin⁡θ$$=ysin$$⁡θ$$+2$$$$π$$$$3=zsin$$⁡θ$$+4$$$$π$$$$3$$ then

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we have, xsin⁡θ$$=ysin$$⁡$$($$θ$$+2$$$$π$$$$3)=zsin$$⁡$$($$θ$$+4$$$$π$$$$3)$$ ⇒$$sin$$⁡θ$$1x=$$$$sin$$⁡$$($$θ$$+2$$$$π$$$$3)1y=$$$$sin$$⁡$$($$θ$$+4$$$$π$$$$3)1z$$ ⇒$$sin$$⁡θ$$1x=$$$$sin$$⁡$$($$θ$$+2$$$$π$$$$3)1y=$$$$sin$$⁡$$($$θ$$+4$$$$π$$$$3)1z$$ $$=$$$$sin$$⁡θ$$+$$$$sin$$⁡$$($$θ$$+2$$$$π$$$$3)+$$$$sin$$⁡$$($$θ$$+4$$$$π$$$$3)1x+1y+1z$$ ⇒$$sin$$⁡θ$$1x=$$$$sin$$⁡$$($$θ$$+2$$$$π$$$$3)1y=$$$$sin$$⁡$$($$θ$$+4$$$$π$$$$3)1z$$ $$=$$$$sin$$⁡θ$$+2sin$$⁡$$($$π$$+$$θ$$)$$$$cos$$⁡π$$31x+1y+1z$$ ⇒ $$sin$$⁡θ$$1x$$×$$(1x+1y+1z)=0$$ ⇒ $$1x+1y+1z=0$$⇒$$xy+yz+zx=0$$

If $xsin θ = ysin (θ + \frac{2 π}{3}) = zsin (θ + \frac{4 π}{3})$ then

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If xsin⁡θ=ysin⁡θ+2π3=zsin⁡θ+4π3 then

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If $xsin θ = ysin (θ + \frac{2 π}{3}) = zsin (θ + \frac{4 π}{3})$ then