The two points at a distance of 3 units from C with the same y coordinate are (-3, -5) and (-9, -5).
Step-by-step explanation:
Step 1:
First, we must plot the point C on the graph. Point C is on the third quadrant so the x value and the y value are negative.
The coordinate of C is (-6, -5).
We need to plot two points that are 3 units away from point C but have the same y coordinate i.e. y = -5.
Step 2:
So assume the two other points are A and B.
The y coordinates of A and B are -5. Only the x coordinate varies. So in order to get the coordinates of point A and B, we add and subtract 3 from the x coordinate of point C.
Point A = (-6+3, -5) = (-3,-5).
Point B = (-6-3, -5) = (-9, -5).
So two points at a distance of 3 units from C with the same y coordinate are (-3, -5) and (-9, -5).
Answer:
Five times the sum of two times a number plus seven all minus three
Step-by-step explanation:
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)
