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3 years ago

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The tee for the sixth hole on a golf course is 400 yards from the tee. On that hole, Marsha hooked her ball to the left, as sket

ched below. Find the distance between Marsha’s ball and the hole to the nearest tenth of a yard. Answer any time! :D

1 answer:

Lubov Fominskaja [6]3 years ago

7 0

Answer:

181.8 yd

Step-by-step explanation:

The law of cosines is good for this. It tells you for triangle sides 'a' and 'b' and included angle C, the length of 'c' is given by ...

c^2 = a^2 +b^2 -2ab·cos(C)

For the given geometry, this is ...

c^2 = 400^2 +240^2 -2(400)(240)cos(16°) ≈ 33,037.75

c ≈ √33037.75 ≈ 181.8 . . . yards

Marsha's ball is about 181.8 yards from the hole.

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3 years ago

On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)

xz_007 [3.2K]

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$P = 10 + 2\sqrt{53}$ units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3), X (2, 3),

Y (4, -4), Z (-3, -2)

Required

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$WX = \sqrt{(-3 - 2)^2 + (3 - 3)^2}$

$WX = \sqrt{(-5)^2 + (0)^2}$

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For XY:

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$XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}$

$XY = \sqrt{-2^2 + (3 +4)^2}$

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For YZ:

$(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)$

$YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}$

$YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}$

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$YZ = \sqrt{53}$

For ZW:

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$ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}$

$ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}$

$ZW = \sqrt{0^2 + (-5)^2}$

$ZW = \sqrt{0 + 25}$

$ZW = \sqrt{25}$

$ZW = 5$

The Perimeter (P) is as follows:

$P = WX + XY + YZ + ZW$

$P = 5 + \sqrt{53} + \sqrt{53} + 5$

$P = 5 + 5 + \sqrt{53} + \sqrt{53}$

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