ANGLES of QUADRILATERAL is 360° and of triangles is 180°)Alternate Solution:Sum of angles of star = (n – 4) × 180°, where n = number of sides FORMING starTherefore ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = (5 – 4) × 180° = 180°Also, sum of interior angles of polygon = (n – 2) × 180° = ∠6 + ∠7 + ∠8 + ∠9 + ∠10 = 3 × 180°Total angles = 720°

"> ANGLES of QUADRILATERAL is 360° and of triangles is 180°)Alternate Solution:Sum of angles of star = (n – 4) × 180°, where n = number of sides FORMING starTherefore ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = (5 – 4) × 180° = 180°Also, sum of interior angles of polygon = (n – 2) × 180° = ∠6 + ∠7 + ∠8 + ∠9 + ∠10 = 3 × 180°Total angles = 720°

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In the given figure, what is the value of ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10?

SRMJEEE Analytical Geometry in SRMJEEE 8 months ago

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∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 = (∠5 + ∠6 + ∠10 + ∠9) + (∠3 + ∠1 + ∠8) + (∠2 + ∠7 + ∠4) = Quad ABCD + ΔPQR + ΔSTU = 360° + 180° + 180° = 720°.(Sum of ANGLES of QUADRILATERAL is 360° and of triangles is 180°)Alternate Solution:Sum of angles of star = (n – 4) × 180°, where n = number of sides FORMING starTherefore ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = (5 – 4) × 180° = 180°Also, sum of interior angles of polygon = (n – 2) × 180° = ∠6 + ∠7 + ∠8 + ∠9 + ∠10 = 3 × 180°Total angles = 720°

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Posted on 18 Nov 2024, this text provides information on SRMJEEE related to Analytical Geometry in SRMJEEE. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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