How do you use sinθ=13 to find secθ?
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given:--for better understanding we shall use x in place of theeta.
sinx=13
we know,sinx=p/h=13
& h^2=p^2+b^2
=>1=(p/h)^2+(b/h)^2
=>1-(13)^2=(b/h)^2
=>b/h=√(168i^2)=2i√(42)
i.e;cosx=2i√(42)
& ultimately secx=1/2i√(42)
sinx=13
we know,sinx=p/h=13
& h^2=p^2+b^2
=>1=(p/h)^2+(b/h)^2
=>1-(13)^2=(b/h)^2
=>b/h=√(168i^2)=2i√(42)
i.e;cosx=2i√(42)
& ultimately secx=1/2i√(42)
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