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PLEASE HELP ME FIGURE THIS OUT 1....

PLEASE HELP ME FIGURE THIS OUT 1....-example-1
PLEASE HELP ME FIGURE THIS OUT 1....-example-1
PLEASE HELP ME FIGURE THIS OUT 1....-example-2

1 Answer

4 votes

Answer:
(11)/(6)

Explanation:

30°-60°-90° triangle has corresponding sides of: b - b√3 - 2b

sec²
((\pi)/(3)) = sec² 60°

sec² 60° =
(1)/(cos^(2)60)

=
((hypotenuse)/(adjacent))^(2)

=
((2b)/(b√(3)))^(2)

=
((2)/(√(3)))^(2)

=
(4)/(3)

45°-45°-90° triangle has corresponding sides of: a - a - a√2

sin²
((\pi)/(4)) = sin² 45°

=
((opposite)/(hypotenuse))^(2)

=
((a)/(a√(2)))^(2)

=
((1)/(√(2)))^(2)

=
(1)/(2)

sec²
((\pi)/(3)) + sin²
((\pi)/(4))

=
(4)/(3) +
(1)/(2)

=
(4)/(3)((2)/(2)) +
(1)/(2)((3)/(3))

=
(8+3)/(6)

=
(11)/(6)

*********************************************************************************

(-12, 5)

  • the x-coordinate is the adjacent side = -12
  • the y-coordinate is the opposite side = 5
  • Use Pythagorean Theorem to find the hypotenuse = 13

(-12)² + (5)² = (hypotenuse)²

144 + 25 = (hypotenuse)²

169 = (hypotenuse)²

13 = hypotenuse

sin θ =
(opposite)/(hypotenuse) =
(5)/(13)

cos θ =
(adjacent)/(hypotenuse) =
-(12)/(13)

tan θ =
(opposite)/(adjacent) =
-(5)/(12)

csc θ =
(hypotenuse)/(opposite) =
(13)/(5)

sec θ =
(hypotenuse)/(adjacent) =
-(13)/(12)

cot θ =
(adjacent)/(opposite) =
-(12)/(5)

User Keidakida
by
8.5k points

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