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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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