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Geometry Level 3

A triangle exists such that the ratio of the sides a : b : c a:b:c is 7 : 8 : 9 7:8:9 . Find cos ⁡ A : cos ⁡ B \cos A : \cos B .

14 : 11 14:11 12 : 63 12:63 14 : 63 14:63 12 : 11 12:11

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Skanda Prasad by Skanda Prasad , 20, India

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1 solution

Ayush G Rai
Sep 21, 2016

We can use cosine formula.Let a = 7 x , b = 8 x a=7x,b=8x and c = 9 x . c=9x.
a 2 = b 2 + c 2 − 2 b c . c o s A a^2=b^2+c^2-2bc.cosA
( 7 x ) 2 = ( 8 x ) 2 + ( 9 x ) 2 − 2 × ( 8 x ) × ( 9 x ) × c o s A ⇒ c o s A = 2 3 {(7x)}^2={(8x)}^2+{(9x)}^2-2\times(8x)\times(9x)\times cosA\Rightarrow cosA=\frac{2}{3}
b 2 = a 2 + c 2 − 2 a c . c o s B b^2=a^2+c^2-2ac.cosB
( 8 x ) 2 = ( 7 x ) 2 + ( 9 x ) 2 − 2 × ( 7 x ) × ( 9 x ) × c o s B ⇒ c o s B = 11 21 {(8x)}^2={(7x)}^2+{(9x)}^2-2\times(7x)\times(9x)\times cosB\Rightarrow cosB=\frac{11}{21}
Therefore, c o s A : c o s B = 14 : 11 . cos A:cos B=\boxed{14:11}.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

nice solution

abhishek alva - 4 years, 8 months ago

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