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Exterior angles in a Triangle tennis2007.org Topical synopsis | Geometry overview | MathBits\" Teacher resources Terms of Use contact Person: Donna Roberts  An exterior edge of a triangle is one angle developed by one side of the triangle and also the extension of an adjacent side the the triangle. FACTS: • Every triangle has 6 exterior angles, two at each vertex.• angle 1 with 6 room exterior angles.• notification that the \"outside\" angles that space \"vertical\" to the angle inside the triangle room NOT called exterior angles of a triangle. The measure up of an exterior angle of a triangle is equal to the sum of the measures of the 2 non-adjacent interior angles. (Non-adjacent inner angles may likewise be referred to as remote inner angles.) FACTS: • an exterior ∠ is same to the addition of the two Δ angles no right next to it. 140º = 60º + 80º; 120º = 80º + 40º; 100º = 60º + 40º • one exterior edge is supplementary to its surrounding Δ angle. 140º is supp come 40º • The 2 exterior angles at every vertex are = in measure due to the fact that they room vertical angles. • The exterior angles (taken one at a vertex) always total 360º
Solution: using the Exterior angle Theorem 145 = 80 + x x = 65 Now, if friend forget the Exterior angle Theorem, you can still obtain the prize by noticing the a right angle has been developed at the crest of the 145º angle. See example 2.
Solution: i forgot the Exterior edge Theorem. The angle surrounding to 145º will kind a directly angle along with 145º including to 180º. That angle is 35º. Now use dominance that sum of ∠s in Δ = 180º. 35 + 80 + x = 180 115 + x = 180 x = 65

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Find m∠DBC. Solution:∠BDC is an exterior angle for ΔABD. m∠BDC = 35 + 25 m∠BDC = 60º 180 = m∠DBC + 60 + 60 m∠DBC = 60º