Answer:
The angles of the ΔABC are:
[tex]m\angle A=30, m\angle B=60, m\angle C=90[/tex]
AB= 14 ft
Step-by-step explanation:
Given:
A triangle ABC, with [tex]m\angle A=30[/tex]°
AB = 2MC
M is the mid-point of AB.
Let AB = [tex]2x[/tex]
Therefore, AM = MB = [tex]\frac{AB}{2}=x[/tex]
Also, MC = [tex]\frac{AB}{2}=x[/tex]
∴ AM = MB = MC = [tex]x[/tex]
Now, consider triangle AMC,
∵ AM = MC
∴ [tex]m\angle MAC = \m\angle MCA = 30[/tex]° ( [tex]m\angle A=m\angle MAC[/tex])
Now, exterior angle BMC is given as the sum of opposite interior angles of triangle AMC.
[tex]m\angle BMC=m\angle MAC+m\angle MCA\\m\angle BMC=30+30=60[/tex]
Consider triangle BMC,
∵ MB = MC
∴ [tex]m\angle MBC = m\angle MCB = a(Let)[/tex]
The sum of all interior angles is equal to 180°.
[tex]m\angle BMC+m\angle MBC+m\angle MCB=180\\60+a+a=180\\2a=180-60\\2a=120\\a=\frac{120}{2}=60[/tex]
Therefore, [tex]m\angle B =a = 60[/tex]°
Also, [tex]m\angle C=m\angle MCA+m\angle MCB = 30+60=90[/tex]°
Therefore, the triangle ABC is a special right angled triangle with measures 30° - 60° - 90°.
For a special right angled triangle 30° - 60° - 90°, the hypotenuse is twice the base.
Here, AB is the hypotenuse and BC is the base. So,
[tex]AB=2BC\\AB=2\times 7=14\ ft[/tex]
Therefore, AB = 14 ft.
