Question

Given: △ABC m∠A=60°, m∠C=45° AB=12 Find: The perimeter.

Answer 1

Perimeter ≈ 44.78 units

Explanation:

In ΔABC first draw a median on line AB from vertex A and name that point as M . Now in ΔAMB,

`\sin 30 =(perpendicular)/(hypotenuse)\\ \sin 30=(AM)/(AB)\\(1)/(2)=(AM)/(12)\\AM=6`

Now, By Pythagoras Theorem in ΔAMB ,

`AB^(2)=AM^(2)+ MB^(2)\\144=36+MB^(2)\\MB=6√(3)`

As median divides the line segment in two equal parts so BC =

`2\cdot MB`

BC = 12√3

Now in ΔAMC,

`\sin 30 =(perpendicular)/(hypotenuse)\\ \sin 30=(AM)/(CA)\\(1)/(2)=(6)/(CA)\\CA=12`

So, perimeter = AB+BC+CA

= 12+12+12√3 = 44.78

Answer 2

Create line segment BM so that it is perpendicular to AC (creating a 90° angle). Now you have created ΔAMB which is a 30°-60°-90° triangle and ΔCMB which is a 45°-45°-90° triangle.

ΔAMB

hypotenuse: AB = 12 (given) = 2x ⇒ x = 6

60°: BM = x√3 ⇒ BM = 6√3

30°: AM = x ⇒ AM = 6

ΔCMB

45°: BM = 6√3 (solved from ΔAMB) = x

45°: MC = x ⇒ MC = 6√3

hypotenuse: BC = x√2 ⇒ BC = 6√6

Perimeter (P)

P = AB + BC + AC (AC = AM + MC per segment addition postulate)

P = 12 + 6√6 + 6 + 6√3

P = 18 + 6√6 + 6√3