Question

in triangle abc angle C is 30 degree more than angle A but angle A is 15 degree less than Angle B find all the three angles​

Answer 1

Answer:

B = 60, A = 45, C=75

Step-by-step explanation:

A = (B-15)

C = A+30 = (B-15) + 30 = B+30

B = B

Sum of All angles is 180

A + B + C = 180

(B-15) + B + (B+30) = 180

3B = 180

B = 60

A = (60-15) = 45

C = (A+30) = 45+30 = 75

Answer 2

$\large \bf \red{ \underline { \underline{Given:-}}}$

• $\sf \angle C \: is \: 30 \degree \:more \: than \: \angle A$

• $\sf \angle A \: is \: 15 \degree \:less\: than \: \angle B$

$\large \bf \blue{ \underline { \underline{To\:Find:-}}}$

• All the three angels of the traingle.

$\large \bf \green{ \underline { \underline{Solution:-}}}$

$\sf let \: \angle B \: be \: x \degree$

∠A is 15 less than ∠B

Then ∠A = (x - 15)

∠C is 30 more than ∠A

∠C = (x - 15 + 30) = ( x + 15)

As we know that sum of all angles in a triangle = 180°

$\sf \rightarrow \: \angle A + \angle B + \angle C = 180 \degree$

$\sf \rightarrow \: (x - 15) + x+ (x + 15)= 180 \degree$

$\sf \rightarrow \: x - 15 + x+ x + 15= 180 \degree$

$\sf \rightarrow \: 3x= 180 \degree$

$\sf \rightarrow \: x= \dfrac{180}{3}$

$\sf \rightarrow \: x= 60 \degree$

Therefore:-

∠C = x = 60°

∠A = x-15 = 60 - 15 = 45°

∠B = x+15 = 60+15 = 75°

Hence;

∠A = 45°

∠B = 75°

∠C = 60°