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 acuteangled:
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 obtuseangled:
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 rightangled
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
From Area of a Triangle to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.