Just two numbers

Geometry Level 3

If A B = 10 AB=10 and cot ⁡ ( α ) + cot ⁡ ( β ) = 3 \cot(\alpha)+\cot(\beta)=3 , what is the area of triangle A B C ABC ?


The answer is 16.667.

Marta Reece by Marta Reece , 69, USA

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2 solutions

Ossama Ismail
May 23, 2018

4 Helpful 0 Interesting 1 Brilliant 0 Confused

Elegant. Thank you.

Marta Reece - 3 years ago

Yes, it's a simple -n- sweet solution, Ossama….thanks!

tom engelsman - 1 year, 2 months ago

You are welcome !

Ossama Ismail - 1 year, 2 months ago
Marta Reece
May 22, 2018

A = a c 2 ⋅ s i n ( β ) = c 2 ⋅ s i n ( β ) ⋅ s i n ( α ) ⋅ c s i n ( γ ) = c 2 2 ⋅ s i n ( α ) ⋅ s i n ( β ) s i n ( γ ) = c 2 2 ⋅ s i n ( α ) ⋅ s i n ( β ) s i n ( 18 0 ∘ − α − β ) = c 2 2 ⋅ s i n ( α ) ⋅ s i n ( β ) s i n ( α + β ) = c 2 2 ⋅ s i n ( α ) ⋅ s i n ( β ) s i n ( α ) ⋅ c o s ( β ) + c o s ( α ) ⋅ s i n ( β ) = c 2 2 ⋅ 1 c o t ( α ) + c o t ) β ) A=\frac{ac}2\cdot sin(\beta)=\frac c2\cdot sin(\beta)\cdot sin(\alpha)\cdot\frac c{sin(\gamma)}=\frac{c^2}2\cdot\frac{sin(\alpha)\cdot sin(\beta)}{sin(\gamma)}=\frac{c^2}2\cdot\frac{sin(\alpha)\cdot sin(\beta)}{sin(180^\circ-\alpha-\beta)}=\frac{c^2}2\cdot\frac{sin(\alpha)\cdot sin(\beta)}{sin(\alpha+\beta)}=\frac{c^2}2\cdot\frac{sin(\alpha)\cdot sin(\beta)}{sin(\alpha)\cdot cos(\beta)+cos(\alpha)\cdot sin(\beta)}=\frac{c^2}2\cdot\frac1{cot(\alpha)+cot)\beta)}

A = 1 0 2 2 ⋅ 1 3 ≈ 16.667 A=\frac{10^2}2\cdot\frac13\approx\boxed{16.667}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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