Question

in triangle ABC, twice of angle A = thrice of angle B = six times of angle C, find the smallest angle of the triangle​

Answer 1

Answer:

let every angle be in term of C

C=C

B = 2C(3B=6C)

A=3C (2A=6C)

so by angle sum property

A+B+C = 180°

3C+2C+C=180

6C=180

C=39

B=60

A=90

Answer 2

Given : In ∆ABC,

$\rm 2\angle A = 3∠B = 6∠C \: \: \: - - - eq ^{n}$

To Find :

The Smallest angke of the Triangle.

Solution :

$\rm from \: \: the \: \: above \: \: equation$

$\rm ∠A = \frac{3∠B}{2}$

and,

$\rm∠C = \frac{∠B}{2}$

then, As we know That,

“ Sum of all Angles in a Triangle is 180°. ”

$\rm⇢ \: \: ∠ A + ∠B + ∠C = 180 \degree$

$\rm⇢ \: \: \frac{3∠B}{2} + ∠B + \frac{∠B}{2} = 180 \degree$

$\rm⇢ \: \: \frac{6∠B}{2} = 180 \degree$

$\rm⇢ \: \: 3∠B = 180 \degree$

$\rm⇢ \: \: ∠B = 60 \degree$

then, Other Angles are :

∠A = 3(60)/2 = 90°

∠B = 60°

∠C = 60/2 = 30°

FINAL ANSWER :

The smallest Angle is ∠C measures 30° .