Answer:

If angle A is 45 degree, then angle O is 90 degree.

In right angled triangle OBC,

OB = 5 cm

OC = 5 cm

so according to pythagoras theorem

BC =

[tex] \sqrt{5 {}^{2} + 5 {}^{2} } [/tex]

[tex] \sqrt{50} [/tex]

[tex]5 \sqrt{2} cm[/tex]

Verified answer

Answer:

[tex]\qquad\qquad\qquad\boxed{ \sf{ \: \bf \: BC = 5 \sqrt{2} \: cm \: }} \\ \\ [/tex]

Step-by-step explanation:

Given that, A triangle ABC is inscribed in a circle with centre O and radius 5 cm.

Further given that, [tex]\angle [/tex]A = 45°

Construction :- Join OB and OC.

We know, angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by the same arc.

So, [tex]\angle [/tex]BOC = 2[tex]\angle [/tex]BAC

[tex]\sf\implies [/tex] [tex]\angle [/tex]BOC = 90°

So, [tex]\sf\implies [/tex] [tex]\triangle [/tex]BOC is right angle triangle.

Now, In right-angle triangle BOC

[tex]\sf \: BO = OC = 5 \: cm \\ \\ [/tex]

By using Pythagoras Theorem, we have

[tex]\sf \: {BC}^{2} = {BO}^{2} + {OC}^{2} \\ \\ [/tex]

[tex]\sf \: {BC}^{2} = {5}^{2} + {5}^{2} \\ \\ [/tex]

[tex]\sf \: {BC}^{2} = 25 + 25 \\ \\ [/tex]

[tex]\sf \: {BC}^{2} = 50 \\ \\ [/tex]

[tex]\sf \: BC = \sqrt{50} \\ \\ [/tex]

[tex]\sf \: BC = \sqrt{5 \times 5 \times 2} \\ \\ [/tex]

[tex]\sf\implies \sf \: BC = 5 \sqrt{2} \: cm \\ \\ [/tex]