(30PTS) What are the slopes of the asymptotes of a hyperbola with equation x^2/2^2-y^2/6^2=1
2 answers:
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Answer:
see explanation
Step-by-step explanation:
Using the cosine and tangent ratios in the right triangle
cos41° = =
=
Multiply both sides by VX
VX × cos41° = 7 ( divide both sides by cos41° )
VX = ≈ 9.3
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tan41° = =
=
Multiply both sides by 7
7 × tan41° = WX, thus
WX ≈ 6.1
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The sum of the 3 angles in a triangle = 180°
Subtract the sum of the given angles from 180° for ∠ X
∠ X = 180° - (90 + 41)° = 180° - 131° = 49°