# Math Help - Word problems...

1. ## Word problems...

Two side of parallelogram form a 120° angle. The longer side of the parallelogram is 18cm long and the diagonal opposite the 120° angle is 24cm long. Find the length of the shorter side and the are of the parallelogram?

2. Start by using the sine rule.

3. i can't start >.< i don't know the diagram or the sketch of the problem i'm confused>.<

4. The diagonal opposite goes from the top left corner to the bottom right corner.

5. Hello, cutiepie32!

Did you make a sketch?

Two side of parallelogram form a 120° angle.
The longer side of the parallelogram is 18cm long.
The diagonal opposite the 120° angle is 24cm long.
Find: (a) the length of the shorter side
. e . .(b) the area of the parallelogram
Code:
    D o - - - - - - - o C
:\ *             \
: \ θ *           \
:  \     *  24     \
h   \ x     *       \
:    \         *     \
:     \           *   \
:      \ 120°      α * \
o     A o - - - - - - - o B
E              18

We have parallelogram $ABCD$ with: . $\angle A = 120^o,\;\;AB = 18,\;\;BD = 24$
Let: . $\theta = \angle BDA,\;\;\alpha = \angle DBA,\;\;x = DA$

Law of Sines: . $\frac{\sin\theta}{18} \:=\:\frac{\sin120^o}{24} \quad\Rightarrow\quad \sin\theta \:=\:0;541265877$

. . Hence: . $\theta \:\approx\:32.77^o$

. . Then: . $\alpha \;=\;180^o - 120^o - 32.77^o \;=\;27.23^o$

Law of Sines: . $\frac{x}{\sin27.23^o} \:=\:\frac{24}{\sin120^o} \quad\Rightarrow\quad x \:\approx\:12.68\text{ cm}$ .(a)

In right triangle $DEB\!:\;\sin\alpha \,=\,\frac{h}{24} \quad\Rightarrow\quad h \;=\;24\sin27.23^o \:\approx\:10.98$

Therefore: . $\text{Area} \;=\;bh \;=\;(18)(10.98) \;=\; 197.64\text{ cm}^2$ .(b)

6. uhm thx but where did you get the sin=0;541265877?