Answered

If cos Θ = negative 4 over 7, what are the values of sin Θ and tan Θ? sin Θ = ±square root 33 over 7; tan Θ = ±square root 33 over 4 sin Θ = ±3 over 7; tan Θ = ±square root 33 sin Θ = ±33 over 7; tan Θ = ±square root 33 over 7 sin Θ = ±square root 33 over 7; tan Θ = ±square root 33 over 7

Answer :

ukshedrack

Answer:

Answer = Option A;

sinθ = ±[tex]\frac{\sqrt{33} }{7}[/tex]

tanθ = ±[tex]\frac{\sqrt{33} }{4}[/tex]

Step-by-step explanation:

cosθ = [tex]-\frac{4}{7}[/tex]

Remove the negative sign and evaluate cosθ = 4/7

cosθ = adjacent/hypotenuse.

Please see attached image for the right-angle triangle representation.

Let opp = length of opposite side of the triangle.

Let adj = length of adjacent side of the triangle.

Let hyp = length of hypotenuse side of the triangle

Using Pythagoras theorem,

[tex]hyp^{2}= opp^{2} +adj^{2} ;\\\\7^{2}= opp^{2}+ 4^{2};\\ \\opp^{2} =7^{2}- 4^{2} \\\\opp^{2} = 49-16\\\\opp^{2} = 33;[/tex]

Take the square root of both sides.

[tex]\sqrt{opp^{2} } =[/tex] ±[tex]\sqrt{33}[/tex]

opp = ±[tex]\sqrt{33}[/tex]

sinθ = opposite/hypotenuse

sinθ = ±[tex]\frac{\sqrt{33} }{7}[/tex]

tanθ = opposite/adjacent

tanθ = ±[tex]\frac{\sqrt{33} }{4}[/tex]

${teks-lihat-gambar} ukshedrack

Other Questions