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

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Question 8: Given a parallelogram with sides 8 cm and 10 cm, and the angle between them is $130^\circ$. Let the parallelogram be ABCD, with AB = 8 cm, AD = 10 cm, and $\angle DAB = 130^\circ$. In a parallelogram, consecutive angles are supplementary, so $\angle ABC = 180^\circ - 130^\circ = 50^\circ$.

a) Calculate the diagonals of the parallelogram.

Step 1: Calculate the length of diagonal AC using the cosine rule in $\triangle ABC$. Sides are AB = 8 cm, BC = AD = 10 cm, and $\angle ABC = 50^\circ$. $AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)$ $AC^2 = 8^2 + 10^2 - 2(8)(10)\cos(50^\circ)$ $AC^2 = 64 + 100 - 160(0.6428)$ $AC^2 = 164 - 102.848$ $AC^2 = 61.152$ $AC = \sqrt{61.152} \approx 7.82 \text{ cm}$

Step 2: Calculate the length of diagonal BD using the cosine rule in $\triangle ABD$. 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 $\boxed{7.82 cm}$ and $\boxed{16.34 cm}$.

b) Calculate angle DBC.

Step 3: Calculate $\angle DBC$ using the cosine rule in $\triangle BCD$. 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.

Step 1: Find diagonal BD using the cosine rule in $\triangle ABD$. Sides are AB = 8 cm, AD = 8 cm, and $\angle DAB = 64^\circ$. $BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(\angle DAB)$ $BD^2 = 8^2 + 8^2 - 2(8)(8)\cos(64^\circ)$ $BD^2 = 64 + 64 - 128(0.4384)$ $BD^2 = 128 - 56.0992$ $BD^2 = 71.9008$ $BD = \sqrt{71.9008} \approx 8.48 \text{ cm}$

Step 2: Find diagonal AC. 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 $\approx 8.48 \text{ cm}$ and AC $\approx 13.57 \text{ cm}$.

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.

Step 1: Find diagonal AC using the cosine rule in $\triangle ABC$. 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 $\boxed{37.59 cm}$.

b) Calculate $\angle ACB$.

Step 2: 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 $\boxed{20^\circ}$.

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