Given [b + c 11 = c + a 12 = a + b 13] for a [ ]ABC with usual notation.

Question

Given $${{b + c} \over {11}} = {{c + a} \over {12}} = {{a + b} \over {13}}$$ for a $$\Delta $$ABC with usual notation.

If   $${{\cos A} \over \alpha } = {{\cos B} \over \beta } = {{\cos C} \over \gamma },$$ then the ordered triad ($$\alpha $$, $$\beta $$, $$\gamma $$) has a value :

Accepted Answer

b + c = 11$$\lambda $$, c + a = 12$$\lambda $$,  a + b = 13$$\lambda $$

$$ \Rightarrow $$  a = 7$$\lambda $$, b = 6$$\lambda $$, c = 5$$\lambda $$

(using cosine formula)

cosA = $${1 \over 5},$$ cosB = $${19 \over 35},$$  cosC = $${5 \over 7},$$

$$\alpha $$ : $$\beta $$ : $$\gamma $$ $$ \Rightarrow $$  7 : 19 : 25

Solutions · 1