Question

The angles of a triangle are in the ratio 2:3:7. The length of the smallest side is 15 cms. The radius of the
circumcircle of the triangle in cms is:​

Answer 1

$\large\underline{\sf{Solution-}}$

Let R be radius of circumcircle to triangle.

↝ As it is given that angles of a triangle are in the ratio 2 : 3 : 7.

↝ So, Let us assume a triangle ABC such that

$\rm :\longmapsto\: \angle \: A = 2x$

$\rm :\longmapsto\: \angle \: B = 3x$

$\rm :\longmapsto\: \angle \: C = 7x$

We know,

↝ Sum of all interior angles of a triangle is supplementary.

$\rm :\longmapsto\: \angle \:A + \angle \:B + \angle \:C = 180 \degree$

$\rm :\longmapsto\: 2x + 3x + 7x= 180 \degree$

$\rm :\longmapsto\:12x= 180 \degree$

$\rm :\longmapsto\:x= 15\degree$

↝ Hence, angles of triangle ABC are as

$\rm :\longmapsto\: \angle \: A = 2 \times 15 = 30 \degree$

$\rm :\longmapsto\: \angle \: B = 3 \times 15 = 45 \degree$

$\rm :\longmapsto\: \angle \: C = 7 \times 15 = 105\degree$

Now,

Let sides of a triangle ABC is represented as

↝ AB = c

↝ BC = a

↝ CA = b

Now, Since,

$\rm :\longmapsto\: \angle \: A = 30 \degree \: is \: smallest \: angle \: in \: \triangle \: ABC$

We know,

↝ Side opposite to smallest angle is always smallest.

↝ BC = a = 15 cm

We know,

Sine Law

$\rm :\longmapsto\:\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = 2R$

$\rm :\longmapsto\:\dfrac{a}{sinA}= 2R$

$\rm :\longmapsto\:\dfrac{15}{sin30 \degree}= 2R$

$\rm :\longmapsto\:\dfrac{15}{ \: \: \dfrac{1}{2} \: \: }= 2R$

$\rm :\longmapsto\:2R = 2 \times 15$

$\bf\implies \:R = 15 \: cm$

Additional Information :-

Projection Formula

$\rm :\longmapsto\:a = b \: cosC + c \: cosB$

$\rm :\longmapsto\:b = a \: cosC + c \: cosA$

$\rm :\longmapsto\:c = a \: cosB + b \: cosA$

Cosine Law

$\rm :\longmapsto\:cosA = \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc}$

$\rm :\longmapsto\:cosB = \dfrac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ac}$

$\rm :\longmapsto\:cosC = \dfrac{ {b}^{2} + {a}^{2} - {c}^{2} }{2ab}$