Find the other dive trigonometric functions OF the acute angle A, given that sec A=2 ?

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Find the other dive trigonometric functions OF the acute angle A, given that sec A=2 ?

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Answer 1 · 2018-04-15T06:46:17

Given,secA = 2.
So, cosA = 1/2 = cos 60.
Then, A = 60 degree.
Now, Sin 60 = $\frac{\sqrt{3}}{2}$ $(sqrt 3)/2$ ,tan 60 = $\sqrt{3}$ $sqrt 3$ , csc 60 = $\frac{2}{\sqrt{3}}$ $2/(sqrt3)$ , cot 60 = $\frac{1}{\sqrt{3}}$ $1/(sqrt3)$ .

ANOTHER WAY
We know, secA = $\frac{2}{1} = \frac{h y p o t e ν s e}{b a s e}$ $2/1= (hypotenuse)/(base)$
So, height = $\sqrt{{(h y p o t e ν s e)}^{2} - {(b a s e)}^{2}}$ $sqrt[(hypotenuse)^2-(base)^2]$
So, height = $\sqrt{2^{2} - 1^{2}}$ $sqrt[2^2-1^2]$
$\Rightarrow \sqrt{3}$ $rArr sqrt3$

Answer 2 · 2018-04-15T06:46:17

Given,secA = 2.
So, cosA = 1/2 = cos 60.
Then, A = 60 degree.
Now, Sin 60 = $\frac{\sqrt{3}}{2}$ $(sqrt 3)/2$ ,tan 60 = $\sqrt{3}$ $sqrt 3$ , csc 60 = $\frac{2}{\sqrt{3}}$ $2/(sqrt3)$ , cot 60 = $\frac{1}{\sqrt{3}}$ $1/(sqrt3)$ .

ANOTHER WAY
We know, secA = $\frac{2}{1} = \frac{h y p o t e ν s e}{b a s e}$ $2/1= (hypotenuse)/(base)$
So, height = $\sqrt{{(h y p o t e ν s e)}^{2} - {(b a s e)}^{2}}$ $sqrt[(hypotenuse)^2-(base)^2]$
So, height = $\sqrt{2^{2} - 1^{2}}$ $sqrt[2^2-1^2]$
$\Rightarrow \sqrt{3}$ $rArr sqrt3$
Now sinA = $\frac{h e i g h t}{h y p o t e ν s e} = \frac{\sqrt{3}}{2}$ $(height)/(hypotenuse)=sqrt3/2$
CosA= $\frac{b a s e}{h y p o t e ν s e} = \frac{1}{2}$ $(base)/(hypotenuse)=1/2$ ,
TanA = $\frac{h e i g h t}{b a s e} = \frac{\sqrt{3}}{1}$ $(height)/(base)=(sqrt3)/1$
CscA = 1/sinA = $\frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$ $1/[(sqrt3)/2]= 2/(sqrt3)$ ,
CotA= 1/tanA = $\frac{1}{\sqrt{3}}$ $1/(sqrt3)$

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Find the other dive trigonometric functions OF the acute angle A, given that sec A=2 ?

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