Calculate the diagonals and an angle of a parallelogram given two sides and an angle.

Sides are AB = 8 cm, BC = AD = 10 cm, and $\angle ABC = 50^\circ$.

$$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$$ $$AC^2 = 8^2 + 10^2 - 2(8)(10)\cos(50^\circ)$$ $$AC^2 = 64 + 100 - 160(0.6428)$$ $$AC^2 = 164 - 102.848$$ $$AC^2 = 61.152$$ $$AC = \sqrt{61.152} \approx 7.82 \text{ cm}$$

Sides are AB = 8 cm, AD = 10 cm, and $\angle DAB = 130^\circ$.

$$BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(\angle DAB)$$ $$BD^2 = 8^2 + 10^2 - 2(8)(10)\cos(130^\circ)$$ $$BD^2 = 64 + 100 - 160(-0.6428)$$ $$BD^2 = 164 + 102.848$$ $$BD^2 = 266.848$$ $$BD = \sqrt{266.848} \approx 16.34 \text{ cm}$$

The diagonals are approximately $\boxed{7.82 \text{ cm}}$ and $\boxed{16.34 \text{ cm}}$.

b) Calculate angle DBC.

In $\triangle BCD$, BC = 10 cm, CD = AB = 8 cm, and BD = 16.34 cm.
Let $\angle DBC = \theta$.

$$CD^2 = BC^2 + BD^2 - 2(BC)(BD)\cos(\theta)$$ $$8^2 = 10^2 + (16.34)^2 - 2(10)(16.34)\cos(\theta)$$ $$64 = 100 + 266.9956 - 326.8\cos(\theta)$$ $$64 = 366.9956 - 326.8\cos(\theta)$$ $$326.8\cos(\theta) = 366.9956 - 64$$ $$326.8\cos(\theta) = 302.9956$$ $$\cos(\theta) = \frac{302.9956}{326.8} \approx 0.9271$$ $$\theta = \arccos(0.9271) \approx 21.95^\circ$$

The angle $\angle DBC$ is approximately $\boxed{21.95^\circ}$.

Question 9:
One side of rhombus ABCD is 8 cm in length. If $\angle DAB = 64^\circ$, find BD and AC.
In a rhombus, all sides are equal, so AB = BC = CD = DA = 8 cm.

Sides are AB = 8 cm, AD = 8 cm, and $\angle DAB = 64^\circ$.

$$BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(\angle DAB)$$ $$BD^2 = 8^2 + 8^2 - 2(8)(8)\cos(64^\circ)$$ $$BD^2 = 64 + 64 - 128(0.4384)$$ $$BD^2 = 128 - 56.0992$$ $$BD^2 = 71.9008$$ $$BD = \sqrt{71.9008} \approx 8.48 \text{ cm}$$

In a rhombus, consecutive angles are supplementary, so $\angle ABC = 180^\circ - 64^\circ = 116^\circ$.
Use the cosine rule in $\triangle ABC$.
Sides are AB = 8 cm, BC = 8 cm, and $\angle ABC = 116^\circ$.

$$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$$ $$AC^2 = 8^2 + 8^2 - 2(8)(8)\cos(116^\circ)$$ $$AC^2 = 64 + 64 - 128(-0.4384)$$ $$AC^2 = 128 + 56.0992$$ $$AC^2 = 184.0992$$ $$AC = \sqrt{184.0992} \approx 13.57 \text{ cm}$$

The diagonals are BD $\approx \boxed{8.48 \text{ cm}}$ and AC $\approx \boxed{13.57 \text{ cm}}$.

Question 10:
ABCD is a rhombus. $\angle ABC = 140^\circ$, and AB = 20 cm. Calculate (a) AC and (b) $\angle ACB$.
In a rhombus, all sides are equal, so AB = BC = CD = DA = 20 cm.

a) Calculate AC.

Sides are AB = 20 cm, BC = 20 cm, and $\angle ABC = 140^\circ$.

$$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$$ $$AC^2 = 20^2 + 20^2 - 2(20)(20)\cos(140^\circ)$$ $$AC^2 = 400 + 400 - 800(-0.7660)$$ $$AC^2 = 800 + 612.8$$ $$AC^2 = 1412.8$$ $$AC = \sqrt{1412.8} \approx 37.59 \text{ cm}$$

The length of AC is approximately $\boxed{37.59 \text{ cm}}$.

b) Calculate $\angle ACB$.

In $\triangle ABC$, AB = BC = 20 cm, so it is an isosceles triangle.
The base angles are equal: $\angle BAC = \angle BCA$.
The sum of angles in a triangle is $180^\circ$.

$$\angle BAC + \angle BCA + \angle ABC = 180^\circ$$ $$2 \times \angle ACB + 140^\circ = 180^\circ$$ $$2 \times \angle ACB = 180^\circ - 140^\circ$$ $$2 \times \angle ACB = 40^\circ$$ $$\angle ACB = \frac{40^\circ}{2} = 20^\circ$$

The angle $\angle ACB$ is $\boxed{20^\circ}$.