Question

The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.​

Answer 1

Step-by-step explanation:

Let B and C be the vertex having exterior angles 104o and 136o respectively.

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136o

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) 

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) 

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) ∠A+∠A+∠B+∠C=240o

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) ∠A+∠A+∠B+∠C=240oAs we know,  ∠A+∠B+∠C=180o 

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) ∠A+∠A+∠B+∠C=240oAs we know,  ∠A+∠B+∠C=180o Hence, ∠A=240o−180o=60o

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) ∠A+∠A+∠B+∠C=240oAs we know,  ∠A+∠B+∠C=180o Hence, ∠A=240o−180o=60oFrom equation (1),

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) ∠A+∠A+∠B+∠C=240oAs we know,  ∠A+∠B+∠C=180o Hence, ∠A=240o−180o=60oFrom equation (1),∠B=104o−60o=44o

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) ∠A+∠A+∠B+∠C=240oAs we know,  ∠A+∠B+∠C=180o Hence, ∠A=240o−180o=60oFrom equation (1),∠B=104o−60o=44oFrom equation (2),

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) ∠A+∠A+∠B+∠C=240oAs we know,  ∠A+∠B+∠C=180o Hence, ∠A=240o−180o=60oFrom equation (1),∠B=104o−60o=44oFrom equation (2),∠C=136o−60o=76o

Let B and C be the vertex having exterior angles 104o and 136o respectively.Ext∠B=104o and Ext∠C=136oUsing theorem "Exterior angle is the sum of interior opposite angles", we get∠A+∠B=104o .......... (1) ∠A+∠C=136o .......... (2)Adding equation (1) and (2) ∠A+∠A+∠B+∠C=240oAs we know,  ∠A+∠B+∠C=180o Hence, ∠A=240o−180o=60oFrom equation (1),∠B=104o−60o=44oFrom equation (2),∠C=136o−60o=76oHence, angles of the triangle are ∠A=60o,∠B=44o and ∠C=76o.