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)
By addition of all the exercise and substracting it from 1 hourJavier runs on treadmill on Monday is 0.40 hour.
Now According to the question javier completed 3 repetitions of each exercise so by multiplication time taken by javier to complete each exercise is as follow;
Squats = 0.8*3= 0.24 hour
Crunches = 0.03*3 = 0.09 hour
Lunges = 0.04*3 = 0.12 hour
curls = 0.05*3 = 0.15 hour
we can calculate total time taken to complete all exercise by Addition is,
0.24 + 0.09 + 0.12 + 0.15
0.60 hour
time taken on treadmill by javier;
1 hour - 0.60 hour
0.40 hour
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Area of a circle = pi*r^2.
Here, A = 380.13 mi. Unfortunately, "mi" is NOT a unit of area.
So I will assume that the problem should read A = 380.13 sq. miles.
380.12 mi^2
Thus, 380.13 mi^2 = 3.14*r^2, and r^2 = ------------------- 121.057 mi^2
3.14
The radius is the sqrt of 121.057 mi^2: 11.00 miles.
The diam. is twice that, or d = 22.00 miles.
Answer:
50 degrees
Step-by-step explanation:
