If $b + c : c + a : a + b = 11 : 12 : 13$ , then $c o s A : c o s B : c o s C =$

7 : 9 : 11

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14 : 11 : 6

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7:19: 25

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8 : 6 : 5

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Solution

The correct option is C 7:19: 25
$b + c = 11 K$ .....(1)
$c + a = 12 K$ ......(2)
$a + b = 13 K$ .....(3)
$a - b = K$ (from (2) - (1) ) .....(4)
$a = 7 K$ (from (3) + (4) ) ......(5)
$b = 6 K$ (from (3) - (5) ) ....(6)
$c = 5 K$ (from (1) - (6) )
Now,
$cos A : cos B : cos C$
$= \frac{b^{2} + c^{2} - a^{2}}{2 b c} : \frac{a^{2} + c^{2} - b^{2}}{2 a c} : \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
$= \frac{6^{2} + 5^{2} - 7^{2}}{2 \times 6 \times 5} : \frac{7^{2} + 5^{2} - 6^{2}}{2 \times 7 \times 5} : \frac{7^{2} + 6^{2} - 5^{2}}{2 \times 7 \times 6}$
$= \frac{12}{60} : \frac{38}{70} : \frac{60}{12 \times 7}$
$= \frac{1}{5} : \frac{19}{35} : \frac{5}{7}$
$= 7 : 19 : 25$

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Similar questions

A B C, \frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}

cos A : cos B : cos C

Fill in the blanks.

(i) $(\frac{- 3}{17}) + (\frac{- 12}{5}) = (\frac{- 12}{5}) + (. . . . .)$

(ii) $- 9 + \frac{- 21}{8} = (. . . . . .) + (- 9)$

(iii) $(\frac{- 8}{13} + \frac{3}{7}) + (\frac{- 13}{4}) = (. . . . . .) + [\frac{3}{7} + (\frac{- 13}{4})]$

(iv) $- 12 + (\frac{7}{12} + \frac{- 9}{11}) = (- 12 + \frac{7}{12}) + (. . . . .)$

(v) $\frac{19}{- 5} + (\frac{- 3}{11} + \frac{- 7}{8}) = {\frac{19}{- 5} + (. . . . .)} + \frac{- 7}{8}$

(vi) $\frac{- 16}{7} + . . . . . = . . . . . . + \frac{- 16}{7} = \frac{- 16}{7}$

1, 3, 5, 7, 9, 11, 13

3, 6, 9, 12, 15

A = {2, 3, 5, 7, 11, 13}

B = {5, 7, 9, 11, 15}

U = {2, 3, 5, 7, 9, 11, 13, 15}

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