Question

In a triangle ABC , if angle A +angle B = 108 degree ,angle B +angle C =130 degree find angle A , angle B , and angle C .

Answer 1

Answer :

• ∠A = 50°

• ∠B = 58°

• ∠C = 72°

Given :

• ∠A + ∠B = 108°

• ∠B + ∠C = 130°

To Find :

• The value of ∠A, ∠B and ∠C.

Step-by-step explanation :

We have ∠A + ∠B = 108° and ∠B + ∠C = 130°

On adding we get,

⟹ (∠A + ∠B) + (∠B + ∠C) = 108 + 130

⟹ ∠A + ∠B + ∠B + ∠C = 238

⟹ (∠A + ∠B + ∠C) + ∠B = 238

⟹ 180 + ∠B = 238

⟹ ∠B = 238 - 180

⟹ ∠B = 58

•°• ∠B = 58°

Now, ∠A + ∠B = 108

⟹ ∠A = 108 - ∠B

= 108 - 58

= 50°

Also, ∠B + ∠C = 130

⟹ ∠C = 130 - ∠B

= 130 - 58

= 72

So, ∠A = 50°

∠B = 58°

∠C = 72°

Answer 2

Answer:

[Refer to attachment ]

$\bold{\angle A = 50\degree}$

$\bold{\angle B =58\degree}$

$\bold{\angle C =72\degree}$

Step-by-step explanation:

Given:-

• $\bold {\angle A + \angle B =108\degree}$
• $\bold{\angle B + \angle C = 130 \degree}$
• $\bold{Value\:of\: \angle A ,\angle B,\angle C=?}$

$\sf \angle A +\angle B = 108\degree .......(i)\\\\\sf \angle B +\angle C = 130\degree.......(ii)$

$\sf At \:first, \angle A + \angle B = 108 \\\\\sf *We\:know, \angle A + \angle B + \angle C = 108\degree \\\\\implies\sf 108 \degree + \angle C = 180\degree \\\\\implies\sf \angle C =180\degree - 108\degree \\\\\implies{\boxed{\sf{\angle C = 72\degree}}}$

$\tt *Putting\:value\:of\:\angle C \:in \:(ii):-$

$\implies\sf \angle B + 72\degree = 130\degree \\\\\implies\sf \angle B = 130\degree - 72\degree \\\\\implies{\boxed{\sf{\angle B = 58\degree}}}$

$\sf *Putting\:value\:of\:\angle B \:in\:(i):-$

$\implies\sf \angle A + 58\degree = 108\degree \\\\\implies\sf \angle A = 108\degree - 58\degree \\\\\implies{\boxed{\sf{\angle A = 50\degree}}}$