If ∆ be the area of a triangle ABC, Proved that, ∆ = ½ bcsin A = ½ ca sin B = ½ ab sin C
That is,
(i) ∆ = ½ bc sin A
(ii) ∆ = ½ ca sin B
(iii) ∆ = ½ ab sin C
Proof:
(i) ∆ = ½ bc sin A
Let ABC is a triangle. Then the following three cases arise:
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Sinac Now khung the above diagram we have, sin C = AD/AC sin C = AD/b, AD = b sin C ………………………..(1) Therefore, ∆ = areaof triangle ABC = một nửa base × altitude |
= ½ ∙ BC ∙ AD = ½ ∙ a ∙ b sin C, = ½ ab sin C Case II: When the triangle ABC is obtuse-angled:
Now size the above diagram we have, sin (180° - C) = AD/AC sin C = AD/AC, sin C = AD/b, AD = b sin C ……………………….. (2) Therefore, ∆ = area of the triangle ABC |
= ½ base x altitude = ½ ∙ BC ∙ AD = ½ ∙ a ∙ b sin C, = ½ ab sin C
Case III: When the triangle ABC is right-angled
Now size the above diagram we have, ∆ = area of triangle ABC = ½ base x altitude = ½ ∙ BC ∙ AD = ½ ∙ BC ∙ AC = ½ ∙ a ∙ b |
= ½ ∙ a ∙ b ∙ 1, = ½ ab sin C Therefore, in all three cases, we have ∆ = ½ ab sin C In a similar manner we can prove the other results, (ii) ∆ =½ ca sin B and (iii) ∆ = ½ ab sin C. ● Properties of Triangles
11 and 12 Grade Math From Area of a Triangle to trang chủ PAGE
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