4 1/3, 4 1/2, 4 5/6, 5 1/6, 5 2/3, 5 6/6
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)
Answer:
9, 32 and 36
Step-by-step explanation:
The perimeter of a triangle is the sum of all the sides. Since you know the perimeter and the expressions for each of the sides, you can set up an equation to solve for the variable and find the values of the other sides:
one side: '4 times the shortest side' = 4s
shortest side: 's'
third side: '23 more than the shortest side' = s + 23
perimeter = 77
Equation: 4s + s + (s + 23) = 77
Combine like terms: 6s + 23 = 77
Subtract 23 from both sides: 6s + 23 - 23 = 77 - 23 or 6s = 54
Divide by 6: 6s/6 = 54/6 or s = 9
one side: 4s = 4(9) = 36
shortest side: s = 9
third side: s + 23 = 9 + 23 = 32
Answer:
Yes
Step-by-step explanation:
1.5(2^x)(2^1)
1.5x2^x x2
3x2^x