Question

The base of the parallelogram is 154 cm and its diagonal make an angle 28° and 43° with the base. Find the length of the longer diagonal. 

Answer 1

we have triangle ABC,

where AB = 154 cm is base,

BC, AC are halves of diagonals,

$\angle BAC=\alpha=28\degree,\angle ABC=\beta=43\degree$

$\angle ACB=\gamma=180\degree-28\degree-43\degree=109\degree$

Then:

$\frac{sin\alpha}{BC}=\frac{sin\beta}{AC}=\frac{sin\gamma}{AB}$

$AC=ABsin\beta/sin\gamma=154sin43\degree/sin109\degree=111.08$ cm

$BC=ABsin\alpha/sin\gamma=154sin28\degree/sin109\degree=76.46$ cm

So, the length of the longer diagonal:

$2AC=2\cdot111.08=222.16$ cm