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
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.
Nov 30, 23 10:59 PM
Nov 30, 23 01:08 PM
Nov 30, 23 01:26 AM
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.