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Anton [14]
3 years ago
8

Need Help, I do not understand.

Mathematics
1 answer:
Lady_Fox [76]3 years ago
8 0

Answer:

11. x=8, y=30

12. w=63°, x=86°, y=38°, z=25°

13. x=18, y=15

Step-by-step explanation:

You have been studying the angles related to transversals and parallel lines, so you know alternate (interior or exterior) angles are congruent (equal) and same-side (interior or exterior) angles are supplementary. You also know that linear angles sum to 180°, the measure of an angle formed by a straight line.

The red arrows in these diagrams signify that the lines so marked are parallel, so the above relationships apply.

11. The diagonal line is a transversal of the top and bottom parallel lines, so the angle marked 5x° is equal to the angle marked 40°. Writing that relationship as an equation, you have ...

... 5x° = 40° . . . . . . . the relationship of the angles

... x = 40/5 = 8 . . . . divide the equation by 5° to find x

The angle marked 3y° and the angle marked with a right-angle symbol are same-side interior angles formed by the vertical transversal with the top and bottom horizontal lines. Hence those angles are supplementary.

... 3y° +90° = 180°

... 3y° = 90° . . . . . . . subtract 90°

... y = 90/3 = 30 . . . .divide by 3°

12. These problems make use of the sum of angles that form a line (180°) and the sum of angles in a triangle (180°).

Together 117° and w form a line, so ...

... 117° + w = 180°

... w = 63° . . . . . . . . . . subtract 117°

The angles of a triangle add to give 180°, so ...

... w + 31° + x = 180°

... 63° +31° + x = 180° . . . . . put in the known value for w

... x = 180° -63° -31°

... x = 86°

Together, x and 56° and y form a line, so the sum of their measures is 180°.

... x + 56° + y = 180°

... 86° + 56° + y = 180° . . . . put in the known value for x

... y = 180° -86° -56° . . . . . . subtract the constants on the left

... y = 38°

The sum of angles y, 117°, and z is 180°.

... 38° + 117° + z = 180° . . . . use the known value for y

... z = 180° -38° -117° . . . . . . subtract the constants on the left

... z = 25°

13. The relationships used in problem 11 also can be used here.

... 5x° + 90° = 180°

... x = (180° -90°)/5°

... x = 18

The angles involving y are same-side interior angles, so are supplementary.

... 5(y +11)° + (4y -10)° = 180°

... 5y +55 + 4y -10 = 180 . . . . . . eliminate parentheses, divide by °

... 9y +45 = 180 . . . . . . . . . . . . . collect terms

... y + 5 = 20 . . . . . . . . . . . . . . . divide by 9

... y = 15 . . . . . . . . . . . . . . . . . . . .subtract 5

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Step-by-step explanation:

Given that;

r = 3

P₁ ( 3, 0°, 30° ),   P₂ ( 3, 45°, 45° )

a)

P.V of P₁

x = rcos∅sin∅ = 3(cos0°) ( sin30°) = (3 × 1 × 0.5) = 1.5

y = rsin∅sin∅  = 3(sin0°) (sin30°)   = (3 × 0 × 0.5) = 0

z = rcos∅        = 3(cos30°)             = ( 3 × 0.866)  = 2.6

∴ P.V  of is OP₁ = [ 1.5i + 0j + 2.6k ]

P.V of P₂

x = rcos∅sin∅ = 3(cos45°) ( sin45°) = (3 × 0.7071 × 0.7071) = 1.5

y = rsin∅sin∅  = 3(sin45°) (sin45°)   = (3 × 0.7071 × 0.7071) = 1.5

z = rcos∅        = 3(cos45°)                 = ( 3 × 0.7071)            = 2.12

∴ P.V  of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b)

Vector connecting P₁ to P₂ is given by

OP₂ - OP₁ = [ 1.5i + 1.5j + 2.12k ] - [ 1.5i + 0j + 2.6k ]

= [ 0i + 1.5j + 0.48k ]  

c)

P₁P₂ → = [ 0i + 1.5j + 0.48k ]  = [ 1.5j + 0.48k ]  

so in a cylindrical coordinate, it should be

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∅ = tan⁻¹[y/π] = π/2

z = 0.48

cylindrical coordinates are (1.5, π/2, 0.48)

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