From the topic:areas and volumes
1 answer:
You might be interested in
Sin ø = opposite/hypotenuse = 2/3, cos ø = adjacent/hypotenuse = ?/3, tan ø = opposite/adjacent = 2/?, sec ø = 1/cos ø
to solve ?, Pythagorean theorem must be applied
hypotenuse^2 = adjacent^2 + opposite^2
Manipulating the equation to find the adjacent value = ?
adjacent = sqrt(hypotenuse^2 - opposite^2) = sqrt(9-4)
adjacent = sqrt(5)
so cos ø = sqrt(5)/3, tan ø = 2/sqrt(5) and sec <span>ø = 3/sqrt(5) since the value is positive the possible equivalent trigonometric function should be positive, the answer should be b </span>
to solve ?, Pythagorean theorem must be applied
hypotenuse^2 = adjacent^2 + opposite^2
Manipulating the equation to find the adjacent value = ?
adjacent = sqrt(hypotenuse^2 - opposite^2) = sqrt(9-4)
adjacent = sqrt(5)
so cos ø = sqrt(5)/3, tan ø = 2/sqrt(5) and sec <span>ø = 3/sqrt(5) since the value is positive the possible equivalent trigonometric function should be positive, the answer should be b </span>