Question

If sin thita= 0.6, find sec thita+ tan thita.

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Answer 1

Step-by-step explanation:

Step-by-step explanation:if sin thita = 0.6

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneous

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64 = 8

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64 = 8 therefore sec thita = 10/8= h/b

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64 = 8 therefore sec thita = 10/8= h/b and tan thita = 6/8= p/b

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64 = 8 therefore sec thita = 10/8= h/b and tan thita = 6/8= p/b hence,

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64 = 8 therefore sec thita = 10/8= h/b and tan thita = 6/8= p/b hence, =>sec thita + tan thita = 10/8+6/8

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64 = 8 therefore sec thita = 10/8= h/b and tan thita = 6/8= p/b hence, =>sec thita + tan thita = 10/8+6/8 = 16/8

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64 = 8 therefore sec thita = 10/8= h/b and tan thita = 6/8= p/b hence, =>sec thita + tan thita = 10/8+6/8 = 16/8 =2

Step-by-step explanation:if sin thita = 0.6 then we know that sin thita= 6/10 = perpendicular/hypotaneousthen By Pythagoras theorom =>(hypotaneous)^2= (perpendicular)^2+ ( base)^2=>(10)^2= (6)^2+ ( base)^2=> ( base)^2= 100- 36 => ( base)^2= 64=>base = √64 = 8 therefore sec thita = 10/8= h/b and tan thita = 6/8= p/b hence, =>sec thita + tan thita = 10/8+6/8 = 16/8 =2the answer is 2