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Please help solve this !!!

Please help solve this !!!-example-1

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Explanation:

3.

remember the law of sine :

a/sin(A) = b/sin(B) = c/sin(C)

a, b, c are the sides. A, B, C are the corresponding opposite angles.

and again, the sum of all angles in a triangle is always 180°.

so, the angle opposite of W = 180 - 90 - 60 = 30°.

and we get

6×sqrt(3)/sin(60) = W/sin(30) = W / 1/2 = 2W

W = 6/2 × sqrt(3)/sin(60) = 3×sqrt(3)/sin(60) =

= 3×sqrt(3) / sqrt(3)/2 = 6

X we get via Pythagoras

X² = (6×sqrt(3))² + W² = 36×3 + 6² = 108+36 = 144

X = sqrt(144) = 12

in the lower triangle the angle opposite of X = 180 - 90 - 45 = 45°

so, it is an isoceles triangle (both legs are equal).

therefore,

Y = X = 12

Z² = 12² + 12² = 144 + 144 = 288

Z = sqrt(288) = sqrt(16×18) = sqrt(16×9×2) =

= 4×3×sqrt(2) = 12×sqrt(2)

4.

x² + 2x - 1 = 2

x² + 2x + 1 = 4

(x + 1)² = 4

x + 1 = ±2

x = ±2 - 1

x1 = 2 - 1 = 1

x2 = -2 - 1 = -3

5.

in a rhombus all 4 sides are equal, and the opposite sides are parallel.

a.

so, PQ = QR

3x + 7 = -x + 17

4x = 10

x = 10/4 = 2.5

b.

the diagonals are perpendicular to each other (90°), intersect each other at their midpoints and cut their corner angles in half.

the angle RSM = angle PSM = 40°.

c.

the sum of all angles in any quadrilateral is always 360°.

in a rhombus the opposing angles are equal.

so,

the angle SPQ = angle SRQ

360 = SPQ + SRQ + PSR + PQR = 2×SPQ + 2×40 + 2×40 =

= 2×SPQ + 160

2×SPQ = 360 - 160 = 200

the angle SPQ = 200/2 = 100°.

User Baiyan Huang
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