If ∆ be the area of a triangle ABC, Proved that, ∆ = ½ bc sin 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:
Case I: When the triangle ABC is acute-angled:
Now form the above diagram we have, sin C = AD/AC
sin C = AD/b, [Since, AC = b] AD = b sin C ……………………….. (1) Therefore, ∆ = area of triangle ABC = 1/2 base × altitude |
= ½ ∙ BC ∙ AD
= ½ ∙ a ∙ b sin C, [From (1)]
= ½ ab sin C
Case II: When the triangle ABC is obtuse-angled:
Now form the above diagram we have,
sin (180° - C) = AD/AC sin C = AD/AC, [Since, sin (π - θ) = sin θ] sin C = AD/b, [Since, AC = b] AD = b sin C ……………………….. (2) Therefore, ∆ = area of the triangle ABC |
= ½ base x altitude
= ½ ∙ BC ∙ AD
= ½ ∙ a ∙ b sin C, [From (1)]
= ½ ab sin C
Case III: When the triangle ABC is right-angled
Now form 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, [Since, ∠C = 90°. Therefore, sin C = sin 90° = 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.
11 and 12 Grade Math
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