Question

in triangle ABC <A=125,BC =8cm find the circum radius of triangle ( sin55=.82)(cos55=.53​

Answer 1

Given :

angle A = 125°

Side BC = 8 cm.

To find :

Circum radius of triangle.

Solution :

Circumradius of a triangle is the radius of a circle in which a triangle can be inscribed.

Lets take angles as

A, B and C

and side AB = c, BC = a and AC = b for our convenience.

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.

Thus, ΔABE∼ΔDBC and

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

which gives

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

(equation 1)

We know that

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

(equation 2)

Putting equation 2 in equation 1, we get

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

so

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

(equation 3)

We know that

a = BC = 8 cm,

A = 125°

so,

sin(A) = sin(125°) = sin (180° - 55°)

sin(A) = sin (55°)

It is given that

sin(55°) = 0.82,

Putting all the values in equation 3, we get

Circum radius of triangle r :

$\bf \: r = \frac{8}{0.82} = </strong><strong>4</strong><strong>.</strong><strong>8</strong><strong>8</strong><strong> \: cm$

so,

circum radius of triangle r = 4.88 cm.

So,

Diameter of circumcircle = 2r = 9.76 cm