Math, asked by happymahi81, 9 months ago

what is the name this formula forarea of triangle 1/2ab sin theta​

Answers

Answered by Anonymous
0

Answer:

side x side x side

Step-by-step explanation:

Answered by aseemalatheef
1

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.

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