# Thread: Length of sides and angle calculations

1. ## Length of sides and angle calculations

Hi,

Can someone tell me how I would find a) the length of side BC and

b) the angle B,on the sketch,

Thanks....

2. Hello, manich44!

As given, the problem has no single solution.
Code:
                D
o
*   *
*       *
*           *
*               *
A o  105.4°           *
*                    o C
*
*
*
*
*
*
o
B

We are given 3 sides and one angle of a quadrilateral.

This does not determine a unique quadrilateral.

3. ## RE

Sorry Guys,
Forgot to include total area = 1276m2

4. Hello, manich44!

I'll give you the Game Plan.
The rest is up to you . . .

Code:
              D
o
34.37  * : α *    37.62
*   :       *
*     :           *
A o 105.4°:               o C
*      :             *
*     :           *
*    :         *
41.4 *   :       *
* θ:     *
* : β *
*: *
o
B

Area of $ABCD \,=\,1276\text{ m}^2$

Find: . $(a)\;BC,\;\;\;(b)\;\angle B$

$\text{Let: }\:\theta = \angle ABD,\;\alpha = \angle BDC,\;\beta = \angle DBC$

[1] Area of $\Delta ABD \,=\,\tfrac{1}{2}(41.4)(34.36)\sin105.4^o \:\approx\:686\text{ m}^2$

. . Hence: ,Area of $\Delta BCD \:=\:1276 - 686 \:=\:590\text{ m}^2$

[2] We know: . $AB = 41.4,\;AD = 34.37,\;\angle A = 105.4^o$

. . . Use the Law of Cosines to find $BD.$

[3] We know all three sides of $\Delta ABD.$
. . . Use the Law of Cosines to find $\theta$

[4] Area of $\Delta BCD \,=\,590\text{ m}^2$

. . . We have: . $\tfrac{1}{2}(BD)(CD)\sin\alpha \:=\:590 \quad\hdots$ Solve for $\alpha.$

[5] We know: $DB,\:DC,\:\angle\alpha$
. . . Use the Law of Cosines to find side $BC$ .(a)

[6] We know all three sides of $\Delta BCD.$
. . . Use the Law of Cosines to find $\beta$

[7] Hence: . $\angle B \:=\:\theta + \beta$ .(b)

5. ## RE

Hi,
Thanks for the help.
I get length BC = 34.30 and angle B 74.6 deg.

Thanks...