In a triangle ABC- if cotA-21-cotb-24- cotC-215then a-b-c is

The correct option is C 19: 16: 5 cot A2 =√(s(s a)(s b)(s c))=s(s a)s ∴ cot A2: cotB2: cotC2 = #160; s a: s b: s c = 1:4 : 15 a=(s ...

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Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

In a triangle...

Question

In a triangle ABC, if $\frac{c o t (A / 2)}{1} = \frac{c o t (b / 2)}{4} = \frac{c o t (C / 2)}{15}$ then a:b:c is

5 : 16 : 19

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19 : 16 : 5

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16 : 19 : 5

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9 : 6 : 5

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Solution

The correct option is C $19 : 16 : 5$
$c o t \frac{A}{2} = \sqrt{\frac{s (s - a)}{(s - b) (s - c)}} = \frac{s (s - a)}{s}$
$∴ c o t \frac{A}{2} : c o t \frac{B}{2} : c o t \frac{C}{2}$

$= s - a : s - b : s - c = 1 : 4 : 15 a = (s - b) + (s - c) = 19 k, b = (s - c) + (s - a) = 16 k, c = (s - c) + (s - a) = 5 k$

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Similar questions

Q. In a triangle ABC, a:b:c=4:5:6, then the ratio of circumradius to inradius will be .

Q. What is the value of

0 - 1 + 2 - 3 + 4 - 5 + 6 - 7......... + 16 - 17 + 18 - 19 + 20

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Q. Assertion :In any triangle

A B C,

a : b : c = 4 : 5 : 6

then

R : r = 16 : 7

Reason: In any triangle

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\frac{R}{r} = \frac{a b c}{4 s}

source $(i) \frac{8}{9} + \frac{- 11}{6} (i i) 3 + \frac{5}{- 7} (i i i) \frac{1}{- 12} a n d \frac{2}{- 15} (i v) \frac{- 8}{19} + \frac{- 4}{57} (v) \frac{7}{9} + \frac{3}{- 4} (v i) \frac{5}{26} + \frac{11}{- 39} (v i i) \frac{- 16}{9} + \frac{- 5}{12} (v i i i) \frac{- 13}{8} + \frac{5}{36} (i x) 0 + \frac{- 3}{5} (x) 1 + \frac{- 4}{5} (x i) \frac{- 5}{16} + \frac{7}{24}$

Q. Let

A = {1, 4, 7, 10, 13, 16, 19},

B = {1, 2, 3, 4, 5, 6},

C = {2, 8, 14, 20},

then

n [A \times (B^{'} \cap C^{'})^{'}] =

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In a triangle ABC, if cot(A/2)1=cot(b/2)4=cot(C/2)15then a:b:c is

The correct option is C 19:16:5cotA2=√s(s−a)(s−b)(s−c)=s(s−a)s∴cotA2:cotB2:cotC2=s−a:s−b:s−c=1:4:15a=(s−b)+(s−c)=19k,b=(s−c)+(s−a)=16k,c=(s−c)+(s−a)=5k

In a triangle ABC, if $\frac{c o t (A / 2)}{1} = \frac{c o t (b / 2)}{4} = \frac{c o t (C / 2)}{15}$ then a:b:c is