Question

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 1

Answer:

Answer = Option A;

sinθ = ±$\frac{\sqrt{33} }{7}$

tanθ = ±$\frac{\sqrt{33} }{4}$

Step-by-step explanation:

cosθ = $-\frac{4}{7}$

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,

$hyp^{2}= opp^{2} +adj^{2} ;\\\\7^{2}= opp^{2}+ 4^{2};\\ \\opp^{2} =7^{2}- 4^{2} \\\\opp^{2} = 49-16\\\\opp^{2} = 33;$

Take the square root of both sides.

$\sqrt{opp^{2} } =$ ±$\sqrt{33}$

opp = ±$\sqrt{33}$

sinθ = opposite/hypotenuse

sinθ = ±$\frac{\sqrt{33} }{7}$

tanθ = opposite/adjacent

tanθ = ±$\frac{\sqrt{33} }{4}$