Consider an isoceles triangle with base a, vertical angle 20 and lateral side each being b
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Consider any triangle ΔABC.ΔABC.
In right triangle ΔABD,ΔABD,
sinθ=BDABsinθ=BDAB
BD=ABsinθBD=ABsinθ
The area of this triangle is easy right?
ar(ΔABC)=12BD⋅ACar(ΔABC)=12BD⋅AC
ar(ΔABC)=12AB⋅ACsinθar(ΔABC)=12AB⋅ACsinθ
We can generalise this as follows, the area of a triangle with consecutive sides a,ba,b and included angle θθ is 12absinθ.12absinθ.
Apllying this to your question,
A=1220×20×12–√A=1220×20×12
A
In right triangle ΔABD,ΔABD,
sinθ=BDABsinθ=BDAB
BD=ABsinθBD=ABsinθ
The area of this triangle is easy right?
ar(ΔABC)=12BD⋅ACar(ΔABC)=12BD⋅AC
ar(ΔABC)=12AB⋅ACsinθar(ΔABC)=12AB⋅ACsinθ
We can generalise this as follows, the area of a triangle with consecutive sides a,ba,b and included angle θθ is 12absinθ.12absinθ.
Apllying this to your question,
A=1220×20×12–√A=1220×20×12
A
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