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 :
JEE · Math · previous-year question
- A.(19, 7, 25)
- B.(7, 19, 25)correct
- C.(5, 12, 13)
- D.(3, 4, 5)
Answer
B. (7, 19, 25)
Explanation
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
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