Question

Given: △ABC, m∠A=120°, AB=AC=1
Find: The radius of circumscribed circle

Answer 1

In Δ ABC, ∠A=120°, AB=AC=1

To draw a circumscribed circle Draw perpendicular bisectors of any of two sides.The point where these bisectors meet is the center of the circle.Mark the center as O.

Then join OA, OB, and OC.

Taking any one OA,OB and OC as radius draw the circumcircle.

Now, from O Draw OM⊥AB and ON⊥AC.

As chord AB and AC are equal,So OM and ON will also be equal.

The reason being that equal chords are equidistant from the center.

AM=MB=1/2 and AN=NC=1/2  [ perpendicular from the center to the chord bisects the chord.]

In Δ OMA and ΔONA

OM=ON [proved above]

OA is common.

MA=NA=1/2  [proved above]

ΔOMA≅ ONA [SSS]

∴ ∠OAN =∠OAM=60° [ CPCT]

In Δ OAN

$\cos60=\frac {AN}{OA}$

$\frac{1}{2}=\frac{\frac{1}{2}}{OA}$

OA=1

∴ OA=OB=OC=1, which is the radius of given Circumscribed circle.

Answer 2

The radius of the circumscribed circle on the considered triangle is 1 unit.

How to find the radius of the circumscribed circle around a triangle?

The intersection of the perpendicular bisectors of any two sides(or all three sides if you want) is the center of the circumscribed circle for that triangle. The length of that intersection point from any of the vertices of the triangle is the radius of the circumscribed circle of that triangle.

For this case, referring to the figure attached below, we see that the since OD and OE are perpendicular bisector, they divide AB and AC in half, and as they are of 1 unit length, thus, we get length of AD = length of AE = 1/2 units.

Triangles DAO and EAO are symmetric as they have same base, common hypotenuse, thus same perpendicular by Pythagoras theorem, and therefore, by SSS symmetry, they are symmetric.

This is the reason why

$m\angle DAO = m\angle EAO$

And since angle A internally is of 120 degrees, thus, we get:

$\angle DAO + \angle EAO = \angle A = 120^\circ\\2\angle DAO = 120\\\angle DAO = 60^\circ = \angle EAO$

Using the cosine ratio, we get for triangle EAO,

$cos(60^\circ) = \dfrac{|AE|}{|OA|}\\\dfrac{1}{2} = \dfrac{1/2}{|OA|}\\\\|OA|= 1 \: \rm unit$

Thus, the radius of the circumscribed circle on the considered triangle is 1 units.

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