Question

in triangle ABC <A=125°,BC=8cm find the diameter of the circumcircle​
(sin 55=0.82)

Answer 1

Given :

Side BC = 8 cm.

Angle A = 125°

To find :

Circumradius of triangle ABC.

Solution :

When a circle is inscribing a triangle inside it,

then the radius of that circle is called Circumradius.

If we take angles as

A, B and C which are opposite to sides BC, AC and AB respectively.

and

let

side AB = c, BC = a and AC = b .

We can get a formula by the following way.

Let ∠C be an acute angle, BD be an altitude of ΔABC and BE be a diameter of the circumcircle.

sin(A) = sin(125°) = sin (180° - 55°)

sin(A) = sin (55°)

It is given that

sin(55°) = 0.82,

Thus, ΔABE∼ΔDBC and

$\frac{c}{h} = \frac{2 \times r}{a}$

which gives

$\bf \: r = \frac{ac}{2h} = \frac{ac}{2\cdot \frac{2\Delta}{b}} = \frac{abc}{4\Delta } \: \:$

(equation 2)

We know that

$\bf \: \Delta \: = \frac{1}{2} bc \sin(A)$

(equation 3)

when we put equation 3 in equation 2, we get

$\bf \: r = (\frac{a}{4 }) (\frac{bc}{\Delta}) \: \: \:$

$\bf \: r = (\frac{a}{4 }) ( \frac{2}{ \sin(A) } ) \:$

So,

$\bf \: r= \frac{a}{2\sin(A)}$

(equation 4)

Then,

We know that

Angle A = 125°

Side a = BC = 8 cm,

Sin A = 0.82

so,

When we put all the values in equation 4, we get

Circumradius of triangle r :

$\bf \: r = \frac{8}{ 2\times 0.82} = \: 4.88 \: cm \: \:$

So,

The diameter of circumcircle d :

$\bf \: d = 2 \times r = 2 \times (4.88) \\ \bf \: d = 9.76 \: cm$