People, geometry 1) in the triangle abc ab=5 bc=4, and its area is 5√3 cm^2 . find the third side of the triangle, if it is known that the angle b is sharp 2 )in the triangle kme angle k=α(alpha), angle m=β (beta), me=b. find the sides of the triangle and its area.

I use the sin rule to find the area

A=(1/2)a*b*sin(∡ab)

1) A=(1/2)*(AB)*(BC)*sin(∡B)
sin(∡B)=[2*A]/[(AB)*(BC)]

we know that
A=5√3
BC=4
AB=5
then

sin(∡B)=[2*5√3]/[(5)*(4)]=10√3/20=√3/2
(∡B)=arc sin (√3/2)= 60°

 now i use the the Law of Cosines 

c2 = a2 + b2 − 2ab cos(C)

AC²=AB²+BC²-2AB*BC*cos (∡B)

AC²=5²+4²-2*(5)*(4)*cos (60) > 25+16-40*(1/2)=21

AC=√21= 4.58 cms

the answer part 1) is 4.58 cms

2) we know that

a/sinA=b/sin B=c/sinC

and

∡K=α

∡M=β

ME=b

then

b/sin(α)=KE/sin(β)=KM/sin(180-(α+β))

KE=b*sin(β)/sin(α)

A=(1/2)*(ME)*(KE)*sin(180-(α+β))

sin(180-(α+β))=sin(α+β)

A=(1/2)*(b)*(b*sin(β)/sin(α))*sin(α+β)=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

KE/sin(β)=KM/sin(180-(α+β))

KM=(KE/sin(β))*sin(180-(α+β)) > KM=(KE/sin(β))*sin(α+β)

the answers part 2) are

side KE=b*sin(β)/sin(α)
side KM=(KE/sin(β))*sin(α+β)
Area A=[(1/2)*b²*sin(β)/sin(α)]*sin(α+β)

Re-do the question and enter the picture

5 cm

step-by-step explanation:

the triangles are similar. therefore the corresponding sides are in proportion:

              cross multiply

            divide both sides by 32