Question

In ∆ABC if tanA/2=5/6,tanB/2=20/37 then find the value of tanC/2 and proove that a+c=2b

Advanced mathematics
class 10th
written by k.c Sinha

Answer 1

HELLO DEAR,

given that:-

tan(A/2)= (5/6)

tan (B/2)= ,(20/37)

we know that;-

$s = \frac{a + b + c}{2} ......(1)$
$\tan( \frac{A}{2} ) = \frac{ \sqrt{(s - b)(s - c)} }{ \sqrt{s(s - a)} } .............(2) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AND \\ \\ \tan( \frac{B}{2} ) = \frac{ \sqrt{(s - a)(s - c)} }{ \sqrt{s(s - b)} } ............(3)\\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AND \\ \\ \\ \tan( \frac{C}{2} ) = \frac{ \sqrt{(s - a)(s - b)} }{ \sqrt{s(s - c)} } ...............(4)$
multiply --(2) and---(3)

we get,

$\tan( \frac{A}{2} ) \times \tan( \frac{B}{2} ) =\frac{ \sqrt{(s - b)(s - c)} }{ \sqrt{s(s - a)} } \times \frac{ \sqrt{(s - a)(s - c)} }{ \sqrt{s(s - b)}} \\ \tan( \frac{A}{2} ) \times \tan( \frac{B}{2} )= > \frac{(s - c)}{s} = \frac{\frac{a + b + c}{2} - c}{\frac{a + b + c}{2}} ..........using(1) \\ = > \frac{5}{6} \times \frac{20}{37} = \frac{a + b - c}{a + b + c} \\ \\ = > \frac{50}{111} = \frac{a + b - c}{a + b + c} \\ \\ = > 50a + 50b + 50c = 111a + 111b - 111c \\ \\ = > 61a + 61b -1 61c..........(5)$

similarly,

$\frac{\tan( \frac{A}{2} )}{\tan( \frac{B}{2} ) } = \frac{{\frac{ \sqrt{(s - b)(s - c)} }{ \sqrt{s(s - a)}}}}{{\frac{ \sqrt{(s - a)(s - c)} }{ \sqrt{s(s - b)}}}} \\ \\ \\ \\ = > \frac{ \frac{5}{6} }{ \frac{20}{37} } = \frac{(s - b)}{(s - a)} \\ \\ = > \frac{37}{24} = \frac{ \frac{a + b + c}{2} - b }{ \frac{a + b + c}{2} - a } ...........using(1) \\ \\ = > \frac{37}{24} = \frac{(a - b + c)}{ (- a + b + c)} \\ \\ = > - 37a + 37b + 37c = 24a - 24b + 24c \\ = > 61a - 13c - 61b...........(6)$
now solving --(5) and--(6)

we get,

61a -161c +61b=0
61a - 13c - 61b=0
_______________
122a- 174c =0

=>a=174c/122

=> a= 87c/61

=>a/87=c/61------(7)

now subtract the Equation--(5)--and --(6)

61a -161c +61b=0
61a - 13c - 61b=0
(-) (+) (+)
_______________
-148c +122=0

122b = 148c

=> b=148c/122

=> b=74c/61

=> b/74=c/61--------(8)

from --(7) and ----(8)

we get,

a/87 =b/74 = c/61

now, we add

(a+c)= >

{87/74 + 61/74 }b

=>2b

hence

◀️a+c=2b▶️

and

$\tan( \frac{C}{2} ) = \frac{ \sqrt{(s - a)(s - b)} }{ \sqrt{s(s - c)} } \\ = > \frac{ \sqrt{(\frac{a + b + c}{2} - a)( \frac{a + b + c}{2} - b)}}{ \sqrt{\frac{a + b + c}{2}(\frac{a + b + c}{2} - c)}} \\ = > \frac{ \sqrt{( \frac{ - a + b + c}{2} )( \frac{a - b + c}{2} )} }{ \sqrt{ (\frac{a + b + c }{2})( \frac{a + b - c}{2} )} } \\ = > \frac{ \sqrt{ (- a + b + c)(2b - b)} } { \sqrt{(2b + b)(a + b - c)} } \\ = > \frac{ \sqrt{(- a + b + c)(b)} }{ \sqrt{(3b)(a + b - c)} } \\ \\ = > \frac{ \sqrt{( - \frac{87c}{61 } + \frac{74c}{61} + c )(b) } }{ \sqrt{(\frac{87c}{61} + \frac{74c}{61 } - c)(3b)} } \\ \\ = > \frac{ \sqrt{ (- 87c + 74c + 61c)} }{ \sqrt{(87c + 74c - 61c)(3b)} } \\ = > \frac{ \sqrt{ 48c} }{ \sqrt{100 c\times 3}} \\ = > \frac{ \sqrt{16} }{ \sqrt{100 } } \\ \\ = > \tan( \frac{C}{2} ) = \frac{4}{10}$
I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Step-by-step explanation:

tanA/2 ={(s-b)(s-c)/s(s-a)}1/2

tanB/2 ={(s-a)(s-c)/s(s-b)}1/2

now first calculate

tanA/2 * tanB/2 = 5/6 * 20/37

61a -161c + 61b =0 .................1

now calcualte

tanA/2 / tanB/2 = 5/6 ÷ 20/37

put the value and simplify

61a -13c -61b =0 ....................2

now solve 1 and 2

a/87 = b/74 =c/61

calculate a+c = [87/74 + 61/74 ]b

=2b