[tex] {\underline {\boxed {\Large {\bf \gray { {120}^{\circ} } }}}} [/tex]
Given :
• In ∆ JKL , all the three interior angles have equal measures.
To calculate:
• Measure of [tex] \angle [/tex] 1 ( Exterior angle )
Calculation:
Here, we can solve this question in two ways.
- By exterior angle property of ∆.
- By linear pair.
______________________________
Let us calculate the measure of interior angles first.
→ Let the measure of each interior angle be x.
As all the interior angles of the triangle are equal,
[tex] \sf { \dashrightarrow {180}^{\circ} = {x}^{\circ} +{x}^{\circ}+{x}^{\circ} } [/tex]
[ By angle sum property of ∆ ]
[tex] \sf { \dashrightarrow {180}^{\circ} = {3x}^{\circ} } [/tex]
[tex] \sf { \dashrightarrow \dfrac{{180}^{\circ}}{3} = {x}^{\circ} } [/tex]
[tex] \sf\red { \dashrightarrow {60}^{\circ} = {x}^{\circ} } [/tex]
Therefore, measure of each interior angle of the triangle is 60°.
By exterior angle property of ∆ :
As we know that,
- Sum of two opposite interior angles of ∆ = Exterior angle
[tex] \sf { \dashrightarrow \angle 1 = \angle K + \angle L} [/tex]
[tex] \sf { \dashrightarrow \angle 1 = {60}^{\circ} + {60}^{\circ} } [/tex]
[tex] \sf \green { \dashrightarrow \angle 1 = {120}^{\circ} } [/tex]
⇒ Hence, measure of angle 1 is 120°.
By linear pair:
→ [tex]\angle [/tex] 1 + [tex]\angle [/tex] J = 180°
→ [tex]\angle [/tex] 1 + 60° = 180°
→ [tex]\angle [/tex] 1 = 180° - 60°
→[tex]\angle [/tex] 1 = 120°
⇒ Hence, measure of angle 1 is 120°.
