Hi,
Can someone tell me how I would find a) the length of side BC and
b) the angle B,on the sketch,
Thanks....
Hello, manich44!
Is there more information?
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.
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 $\displaystyle ABCD \,=\,1276\text{ m}^2$
Find: .$\displaystyle (a)\;BC,\;\;\;(b)\;\angle B$
$\displaystyle \text{Let: }\:\theta = \angle ABD,\;\alpha = \angle BDC,\;\beta = \angle DBC$
[1] Area of $\displaystyle \Delta ABD \,=\,\tfrac{1}{2}(41.4)(34.36)\sin105.4^o \:\approx\:686\text{ m}^2$
. . Hence: ,Area of $\displaystyle \Delta BCD \:=\:1276 - 686 \:=\:590\text{ m}^2$
[2] We know: .$\displaystyle AB = 41.4,\;AD = 34.37,\;\angle A = 105.4^o$
. . . Use the Law of Cosines to find $\displaystyle BD.$
[3] We know all three sides of $\displaystyle \Delta ABD.$
. . . Use the Law of Cosines to find $\displaystyle \theta$
[4] Area of $\displaystyle \Delta BCD \,=\,590\text{ m}^2$
. . . We have: .$\displaystyle \tfrac{1}{2}(BD)(CD)\sin\alpha \:=\:590 \quad\hdots $ Solve for $\displaystyle \alpha.$
[5] We know: $\displaystyle DB,\:DC,\:\angle\alpha$
. . . Use the Law of Cosines to find side $\displaystyle BC$ .(a)
[6] We know all three sides of $\displaystyle \Delta BCD.$
. . . Use the Law of Cosines to find $\displaystyle \beta$
[7] Hence: .$\displaystyle \angle B \:=\:\theta + \beta$ .(b)