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) areside 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)
Parallelogram have 4 sides.
Use the distributive property. 2 multiply the last and negative 4 in the parentheses.
2l + 2(l - 4) = 72 m
Combine Like terms
2l + 2l - 8 = 72 m
4l - 8 = 72 m
Get l all by itself by adding 8 to both sides
4l - 8 = 72 m
+8. +8
Divide both sides by 4
4l = 80
4. 4
l= 20
The width of the parallelogram is four meters less than its length.
4 sides
2 length = 20
2 width = 16
I got 16 by subtracting 4 from 20
Comment if you need any other help :)
Answer:
Step-by-step explanation:
the 6th power of 10 would be 10^6, then you'd multiply it by 3 which would make it 10^6*3
you could then simplify it and say 10^18 (multiply the exponents [6 and 3])
Answer: 69
Step-by-step explanation:
The two angles are a linear pair, which means that they lie on the same line. Therefore, you just have to substract angle 1 with 180.
180-111=69
