Answer:
x=57
Step-by-step explanation:
63+60= 123
A triangle is 180 degrees
180-123= 57
x=57
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)
Pls. see attachment.
We need to solve for the angles of the smaller triangle in
order to solve for the angle of the larger triangle which would help us solve
the missing measurement of a side.
Given:
51 degrees.
Cut the triangle into two equal sides and it forms a right
triangle. All interior angles of a triangle sums up to 180 degrees.
180 – 51 – 90 = 39 degrees
39 degrees * 2 = 78 degrees.
Angle Q is 78 degrees.
In the bigger triangle, 4.3 is the hypotenuse. We need to
solve for the measurement of the long leg which is the opposite of the 78
degree angle.
We will use the formula:
Sine theta = opposite / hypotenuse
Sin(78 deg) = opposite / 4.3
Sin(78 deg) * 4.3 = opposite
4.21 = opposite. This is also the height of the triangle.
Area of a triangle = ½ * base * height
A = ½ * 3units * 4.21units
A = 6.315 square units.