In â?? ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C

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In Δ ABC, right-angled at B Using Pythagoras theorem AC² = AB² +BC² AC² = 576 + 49 = 625 AC = √625 AC = ±25 Now (i) In a right angle triangle ABC where B=90° , Sin A = BC/AC =7/25 CosA =AB/AC =24/25 (ii) Sin C =AB/AC =24/25 Cos C =BC/AC =7/25
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In Δ ABC, right-angled at B Using Pythagoras theorem AC² = AB²+BC² AC² = 576 + 49 = 625 AC = √625 AC = ±25 Now (i)In a right angle triangle ABC where B=90° , Sin A = BC/AC =7/25 CosA =AB/AC =24/25 (ii) Sin C =AB/AC =24/25 Cos C =BC/AC =7/25 read less
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SinA = bc/ac=7/25 CosA=ab/ac=24/25 SinC =ab/ac=24/25 CosC=bc/ac=7/25
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Δ ABC, right-angled at B Using Pythagoras theorem AC² = AB² +BC², which is right-angled at B. AC² = 576 + 49 = 625 AC = √625 AC = ±25 Now (i) In triangle ABC, Sin A = BC/AC= opposite side ÷hypotenuse =7/25, CosA =AB/AC=adjacent side÷ hypotenuse =24/25 (ii)...
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Δ ABC, right-angled at B Using Pythagoras theorem AC² = AB² +BC², which is right-angled at B. AC² = 576 + 49 = 625 AC = √625 AC = ±25 Now (i) In triangle ABC, Sin A = BC/AC= opposite side ÷hypotenuse =7/25, CosA =AB/AC=adjacent side÷ hypotenuse =24/25 (ii) Sin C =AB/AC =24/25; Cos C =BC/AC=7/25 read less
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