what is the name this formula forarea of triangle 1/2ab sin theta
Answers
Answer:
side x side x side
Step-by-step explanation:
Answer:
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
Area of Acute-angled Triangle
= 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
Area of Obtuse-angled Triangle
= 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
Area of Right-angled Triangle
= 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.
Step-by-step explanation:
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
Area of Acute-angled Triangle
= 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
Area of Obtuse-angled Triangle
= 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
Area of Right-angled Triangle
= 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.