Calculate the diagonals of a parallelogram with sides 8 cm and 10 cm, and the angle between them is 130 degrees.

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 $\angle DAB = 130^\circ$.

$$BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos(\angle DAB)$$ $$BD^2 = 8^2 + 10^2 - 2(8)(10)\cos(130^\circ)$$ $$BD^2 = 64 + 100 - 160(-0.6428)$$ $$$$