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सी.आई.एस.सी.ई.आईसीएसई ICSE Class 8

Given : in Quadrilateral Abcd ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(A+2)° and ∠B = 2(2a+7)°. Calculate ∠A.

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प्रश्न

Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(a+2)° and ∠B = 2(2a+7)°.
Calculate ∠A.

योग
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उत्तर

∵ ∠C = 64° (Given)

∴ ∠D = ∠C – 8° = 64°- 8° = 56°

∠A = 5(a+2)°

∠B = 2(2a+7)°

Now ∠A + ∠B + ∠C + ∠D = 360°

5(a+2)° + 2(2a+7)° + 64° + 56° = 360°

5a + 10 + 4a + 14° + 64° + 56° = 360°

9a + 144° = 360°

9a = 360° – 144°

9a = 216°

a = 24°

∴ ∠A = 5 (a + 2) = 5(24+2) = 130°

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