Answer:
The angles of the ΔABC are:
![m\angle A=30, m\angle B=60, m\angle C=90](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rlfm5rcmu7nkiqwrju90bvd1mkvlkaoflx.png)
AB= 14 ft
Explanation:
Given:
A triangle ABC, with
°
AB = 2MC
M is the mid-point of AB.
Let AB =
![2x](https://img.qammunity.org/2020/formulas/mathematics/high-school/ehg1x3n3vnt1s4e4gae0epm9rgujr7wlmt.png)
Therefore, AM = MB =
![(AB)/(2)=x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ub0rpkph1i6hwegikhga397hyt5ep9hoo5.png)
Also, MC =
![(AB)/(2)=x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ub0rpkph1i6hwegikhga397hyt5ep9hoo5.png)
∴ AM = MB = MC =
![x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k3ozza40nv61jy1offmxaxutrb6y1c3ly5.png)
Now, consider triangle AMC,
∵ AM = MC
∴
° (
)
Now, exterior angle BMC is given as the sum of opposite interior angles of triangle AMC.
![m\angle BMC=m\angle MAC+m\angle MCA\\m\angle BMC=30+30=60](https://img.qammunity.org/2020/formulas/mathematics/middle-school/i3wiya8k2i9ub1jh1ndqd5j3hcyg915lht.png)
Consider triangle BMC,
∵ MB = MC
∴
![m\angle MBC = m\angle MCB = a(Let)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ybty9cq20367mru6817vqs4jcb13lgzu2e.png)
The sum of all interior angles is equal to 180°.
![m\angle BMC+m\angle MBC+m\angle MCB=180\\60+a+a=180\\2a=180-60\\2a=120\\a=(120)/(2)=60](https://img.qammunity.org/2020/formulas/mathematics/middle-school/z3w33mb12zz0iznctx2t4zxn936ixqexpe.png)
Therefore,
°
Also,
°
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,
![AB=2BC\\AB=2* 7=14\ ft](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wpf09uuyni14dcu0mtb67czshhc4w9hefe.png)
Therefore, AB = 14 ft.