Question

Please help me with this algebra question please u need help

Answer 1

Given↷

$\huge\rightarrow\sin( θ ) = \frac{ - 2 \sqrt{5} }{5}$

To find↷

• cos(θ) in IV quadrant

Answer↷

$\huge\rightarrow\cos( θ ) = \frac{ \sqrt{5} }{5}$

Solution↷

We know that,

sine is the ratio of adjacent side over the hypotenuse.

$\rightarrow\sin( θ ) = \frac{ - 2 \sqrt{5} }{5} = \frac{adjacent \: side}{hypotenuse}$

also,

cosine is the ratio of base over the hypotenuse.

$\rightarrow\cos( θ ) = \frac{?}{5}$

• base = √(h²-a²)
• h= hypotenuse
• a= adjacent side

$\rightarrow\cos( θ ) = \frac{ \sqrt{ {5}^{2} - ( - 2\sqrt{ 5} {}^{} ) {}^{2} } }{5}$

$\rightarrow\cos( θ ) = \frac{ \sqrt{ {5}^{2} - {(2\sqrt{ 5})}^{2} } }{5}$

$\rightarrow\cos( θ ) = \frac{ \sqrt{25 - 20} }{5}$

$\rightarrow\cos( θ ) = \frac{ \sqrt{5} }{5}$

since ,cosine is always positive in the quadrant IV, option 3rd =√5/5 is correct ✓

Answer 2

Note:-

★theta to be used as A

$\\ \rm\Rrightarrow sinA=\dfrac{-2\sqrt{5}}{5}$

Now

• cosA=1-sin²A

$\\ \rm\Rrightarrow cosA=\sqrt{1-(dfrac{-2\sqrt{5}}{5})^2}$

$\\ \rm\Rrightarrow cosA=\sqrt{1-\dfrac{20}{25}}$

$\\ \rm\Rrightarrow cosA=\sqrt{\dfrac{25-20}{25}}$

$\\ \rm\Rrightarrow cosA=\dfrac{\sqrt{5}}{5}$

Option C