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.**

**The Law of Sines or The Sine Rule****Theorem on Properties of Triangle****Projection Formulae****Proof of Projection Formulae****The Law of Cosines or The Cosine Rule****Area of a Triangle****Law of Tangents****Properties of Triangle Formulae****Problems on Properties of Triangle**

**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.