WebThe area of the figure formed by joining the mid points If x4+1x4=194 then x3+1x3= 76 52 64 none of these x4+1x In the figure if l m n p and ∠1=850 find ∠2 In th WebQuestion From - NCERT Maths Class 9 Chapter 6 SOLVED EXAMPLES Question – 6 LINES AND ANGLES CBSE, RBSE, UP, MP, BIHAR BOARDQUESTION TEXT:-In Fig. 6.27, AB ...
Example 6 - In figure, AB CD and CD EF. Also EA ⊥ AB - teachoo
WebAC = AB, AB // DF, BC // ED, AC // EF and ∠ CAB = 70°. Find ∠ x, ∠ y and ∠ z. Δ ABC is an isosceles triangle. ∠ ACB = ∠ ABC = (180° − 70°) ÷ 2 = 55° DF // AE and AD // EF So, AEFD is a parallelogram . ∠ ... AB = AC, ∠ AEC = 53° and ... ∠ t = ∠ FCE − ∠ ACE = ∠ FCE − ∠ ACB = 127° − 38 ° = 89 ... Web(1) AB and CD are parallel lines cut by a transversal. So the co-interior angles formed are supplementary. x + y = 180°. Since x = z, We get y + z = 180°......... (2) Let, y = 3a, z = 7a [Since, y : z = 3 : 7] Substituting the values in equation (2), 3a + 7a = 180° 10a = 180° a = 180°/10 a = 18° ∴ y = 3a = 3 × 18 = 54° y = 54° ∴ x + y = 180° emeralandthea gmail.com
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WebMay 1, 2024 · Given : From figure : ∠ECD 35°, ∠ACE = 22° , ∠CEF = 145° . To Prove : AB EF Proof: ∠ACD = ∠ACE + ∠ECD ∠ACD = 22° + 35° ∠ACD = 57° ∴ ∠BAC = ∠ACD Thus lines AB and CD are intersected by the line BC such that ∠ABC = ∠BCD i.e. , the alternate angles are equal. ∴ AB CD ……….. (1) Now, ∠DCE + ∠CEF = 35° + 145° ∠DCE + ∠CEF = 180° WebGiven: In the figure, if AB CD and CD EF, ∠ B A C = 70 0, ∠ C E F = 130 0 ∵ E F C D ∴ ∠ E C D + ∠ C E F = 180 0 (Co-interior angles) ⇒ ∠ E C D + 130 0 = 180 0 ∴ ∠ E C D = 180 0 − 130 … WebSolution Verified by Toppr Correct option is A) It is given that AB∥CD and BC is transversal from the figure we know that ∠BCD and ∠ABC are alternate interior angles so we get … emerail cookware sold at