Question

In a ∆ABC, ∠A + ∠B = 125⁰ and ∠A + ∠C = 115⁰ ,find the values of ∠A,∠B and ∠C respectively.​

Answer 1

Step-by-step explanation:

We're given with :

• ∠A + ∠B = 125°
• ∠A + ∠C = 115°

To get this question's solution, one needs to be aware of the Angle Sum Property of Triangle.

Angle Sum Property of Triangle states that the sum of all angles in a triangle is 180°.

• Value of ∠C

$\longrightarrow$ ∠A + ∠B + ∠C = 180°

We know the value of ∠A + ∠B, substitute it.

$\longrightarrow$ 125° + ∠C = 180°

$\longrightarrow$ ∠C = 180° - 125°

$\longrightarrow$ ∠C = 55°

• Value of ∠A

$\longrightarrow$ ∠A + ∠C = 115°

$\longrightarrow$ ∠A + 55° = 115°

$\longrightarrow$ ∠A = 115° - 55°

$\longrightarrow$ ∠A = 60°

• Value of ∠B

$\longrightarrow$ ∠A + ∠B = 125°

$\longrightarrow$ 60° +∠B = 125°

$\longrightarrow$ ∠B = 125° - 60°

$\longrightarrow$ ∠B = 65°

Therefore, the angles are

• ∠A = 60°
• ∠B = 65°
• ∠C = 55°

Answer 2

Given:-

✭ ∠A + ∠B = 125⁰ ( First equation )

✭ ∠A + ∠C = 115⁰ ( Second equation)

To Find:-

⇝ The values of ∠A , ∠B , ∠C of a ∆ABC

Now Substituting ∠B in terms of ∠A from equation( i ) :

➾ ∠A + ∠B = 125⁰

➾ ∠B = 125⁰ - ∠A

✪ So ∠B = 125⁰ - ∠A

Now Substituting ∠C in terms of ∠A from equation( ii ) :

➾ ∠A + ∠C = 115⁰

➾ ∠C = 115⁰ - ∠A

✪ So ∠C = 115⁰ - ∠A

Now leave ∠A in its own form :

➾ ∠A = ∠A

➾ ∠B = 125⁰ - ∠A

➾ ∠C = 115⁰ - ∠A

Sum of all angles of Triangle = 180⁰

➾ ∠A + ∠B + ∠C = 180⁰

➾ ∠A + ( 125⁰ - ∠A ) + ( 115⁰ - ∠A ) = 180⁰

➾ ∠A + 125⁰ - ∠A + 115⁰ - ∠A = 180⁰

➾ -∠A + 240⁰ = 180⁰

➾ ∠A = 240⁰ - 180⁰

∴ ∠A = 60⁰

So ∠B = 125⁰ - ∠A

➾ ∠B = 125⁰ - 60⁰

∴ ∠B = 65⁰

So ∠C = 115⁰ - ∠A

➾ ∠C = 115⁰ - 80⁰

∴ ∠C = 55⁰

Now :-

✪ ∠A = 60⁰ Ans.

✪ ∠B = 65⁰ Ans.

✪ ∠C = 55⁰ Ans.