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In Δ ABC, if cos A = sin B - cos C then show that it is a rightangled triangle. - Mathematics and Statistics

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प्रश्न

In Δ ABC, if cos A = sin B - cos C then show that it is a right-angled triangle.

बेरीज
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उत्तर

cos A = sin B - cos C

∴ cos A + cos C = sin B

∴ `2 cos (("A + C")/2). cos (("A - C")/2)` = sin B

∴ `2 cos (pi/2 - "B"/2). "cos" (("A - C")/2)` = sin B .....[∵ A + B + C = π]

∴ `2 sin  "B"/2. cos (("A - C")/2) = 2 sin  "B"/2. "cos" "B"/2`

∴ `cos  (("A - C")/2) = cos  "B"/2`

∴ `("A - C")/2 = "B"/2`

∴ A - C = B

∴ A = B + C

∴ A + B + C = 180° gives

∴ A + A = 180°

∴ 2A = 180°

∴ A = 90°

∴ Δ ABC is a right angled triangle.

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Trigonometric Equations and Their Solutions
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ १०९]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 8 | पृष्ठ १०९

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