Chapter 12.8, Problem 18WE

To determine

Find the lengths of the diagonals.

Answer to Problem 18WE

The lengths of the diagonals are about 2.62 and 1.35

Explanation of Solution

Formula Used:

Law of Cosine :

  $a^2=b^2+c^2-2bc\cos A$

Law of Sine:

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

  

Calculation:

Represent the situation and solve :

  

The longer diagonal bisects the shorter diagonal.

So, AO = OC

The diagonals are perpendicular to each other.

Use Law of Cosine in triangle ABD:

  $\begin{array}{l}B{D}^2=A{B}^2+A{D}^2-2(AB)(AD)\cos \angle DAB\\ \Rightarrow B{D}^2=2^2+1^2-2(2)(1)\cos 118°\\ \Rightarrow BD\approx 2.62\end{array}$

Now, use Law of Sine in triangle ABD:

  $\begin{array}{l}\frac{\sin 118°}{2.62}\text{=}\frac{\sin \angle ADB}{2}\\ \Rightarrow \sin \angle ADB=2\cdot \frac{\sin 118°}{2.62}\\ \Rightarrow \angle ADB\approx 42.38°\end{array}$

In triangle AOD,

  $\begin{array}{l}\sin 42.38°=\frac{AO}{1}\\ \Rightarrow \text{AO=OC=0}\text{.674}\\ \Rightarrow \text{AC=2(0}\text{.674) = 1}\text{.348}\\ \end{array}$

So, the lengths of the diagonals are about 2.62 and 1.35