Question

Please solve both
1.In any triangle ABC, if angle A - angle B = 33° and angle B - angle C = 18°, then
find the value of angle B.

2.In any triangle ABC, if 2 angle A = 3 angle B = 6 angle C, then the value of angle A.​

Answer 1

Answer:

l) angle A - angle B = 33°(given)

so, angle A= 33+ angle B. (1)

angle B - angle C = 18°(given)

angle B= 18+angle C

angle C= angle B- 18°. (2)

now, as we know angle sum property of a triangle = 180°

so , angle A+angle B+angle C = 180°

putting (1) and (2)

(33+angle B ) + angle B + (angle B -18) = 180°

3(angle B) +33-18= 180

3(angle B)= 180- 15

angle B= 165/3

angle B = 55°

ll) 2 angle A = 3 angle B = 6 angle C

it means ... 2(angle A)= 3(angle B)

angle B = 2(angle A)/3. (1)

also, 2(angle A)= 6(angle C)

angle C= 2(angle A)/6

angle C= angle A/ 3. (2)

We know angle sum property of triangle = 180°

angle A + angle B + angle C = 180°

putting (1) and (2)

angle A + 2(angle A/3) + angle A/3= 180°

Finding the LCM to make the base same

3(angle A )/3 + 2(angle A)/3 + angle A/3 = 180

6(angle A)/3= 180

2(angle A)= 180

angle A = 180 /2

angle A = 90°

hope it helps

Answer 2

Property Used :-

Sum of the interior angles of a triangle is 180°.

$\large\underline{\sf{Solution-1}}$

Given that,

$\rm :\longmapsto\: \angle \:A \: - \: \angle \:B = 33\degree$

$\rm :\longmapsto\: \angle \:A \: = \: \angle \:B + 33\degree - - (1)$

and

$\rm :\longmapsto\: \angle \:B \: - \: \angle \:C = 18\degree$

$\rm :\longmapsto\: \angle \:B \: - \:18\degree \: = \: \angle \:C - - - (2)$

We know,

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

Please follow the atrachment for further solution.