Question

In ΔABC, angleA+angleB=105 and AngleB+ AngleC=112,then find Angle A, B and C​

Answer 1

angle A + angle B = 105°

angle A = 105° - angle B

angle B + angle C = 112°

angle C = 112° - angle B

In triangle ABC,

angle A + angle B + angle C = 180° (sum of angles of a triangle)

(105°-angle B) + angle B +(112°-angle B) = 180°

105° +112° - angle B + angle A - angle B = 180°

217° -2 angle B + angle B = 180°

-angle B = 180°-217°

-angle B = -37°

angle B = 37°

thus, angle A = 105° - angle B = 105°-37° = 68°

angle C = 112°-angle B = 112°-37° = 75°

Answer 2

Answer:

The Value of ∠A , ∠B, ∠ C is 68°, 37°, 75° respectively.

Step-by-step explanation:

∵Sum of all interior angles of a triangle  = 180°

∴ ∠A + ∠B + ∠C = 180° -----------------------(eq.1)

Given that,

∠A + ∠B = 105° -------------------( eq. 2 )

and,

∠B + ∠C = 112° --------------------( eq. 3)

On adding (eu.2) and (eq.3),

We get,

∠A + 2∠B + ∠C = 217° -------------------------(eq.4)

Now, Subtracting (eq.1) from (eq.4),

We get,

∠A+∠C+ 2∠B - ∠A - ∠C - ∠B = (217°-180°)  

⇒∠B = 37°

∴∠A + ∠B = 105°                         and  ∠B + ∠C = 112°

⇒∠A = 105° - 37°                             ⇒ ∠C = 112° - 37°

⇒ ∠A = 68°                                      ⇒∠C = 75°

∴The required ∠A, ∠B, ∠C are 68°, 37°, 75° respectively.

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