Answer to Question #347659 in Trigonometry for Aeron Jay

Question

Two adjacent sides of a parallelogram meet at an angle of 450 and have lengths of 5 and 11 feet respectively. Find the length of the diagonals.

Answer 1

We can use the Law of Cosines in the triangle ABD:

$BD^2=AB^2+AD^2-2AB\cdot AD cos\angle A,$

$BD^2=5^2+11^2-2\cdot 5\cdot 11 \cdot cos45\degree=$

$=25+121-2\cdot 55\cdot \frac{\sqrt{2}}{2}=146-55\sqrt{2},$

$BD=\sqrt{146-55\sqrt{2}}.$

Let's find the angle B:

$\angle B=180\degree - \angle A=180\degree -45\degree =135\degree.$

And now we can use the Law of Cosines in the triangle ABC:

$AC^2=AB^2+BC^2-2AB\cdot BC cos\angle A,$

$BD^2=5^2+11^2-2\cdot 5\cdot 11 \cdot cos135\degree=$

$=25+121-2\cdot 55\cdot (-\frac{\sqrt{2}}{2})=146+55\sqrt{2},$

$AC=\sqrt{146+55\sqrt{2}}.$

Answer: $\sqrt{146-55\sqrt{2}},$ $\sqrt{146+55\sqrt{2}}.$