Question

if a triangle is given by 19a=b+c,then find the value of cot[b/2]cot[c/2]

Answer 1

Answer:

Given : b+c = 3a

According to the half range formula, Cot\frac { B }{ 2 } =\sqrt { \frac { (s-a)(s-c) }{ s(s-b) } }Cot2B=s(s−b)(s−a)(s−c)

Cot\frac { C }{ 2 } =\sqrt { \frac { (s-a)(s-b) }{ s(s-c) } }Cot2C=s(s−c)(s−a)(s−b)

We know that {\frac{(a+b+c)}{2}} = s2(a+b+c)=s , given that b + c = 3a

                S = {\frac{( a + 3a )}{2}} ={\frac{ 4a}{ 2}} = 2aS=2(a+3a)=24a=2a

Cot\frac { B }{ 2 } Cot\frac { C }{ 2 } = \sqrt { \frac { (s-a)(s-c) }{s(s-b) } } \times \sqrt { \frac { (s-a)(s-b) }{ s(s-c) } }Cot2BCot2C=s(s−b)(s−a)(s−c)×s(s−c)(s−a)(s−b)

              = \sqrt { \frac { (2a-a)(2a-c) }{ 2a(2a-b) } } \times \quad \sqrt { \frac { (2a-a)(2a-b) }{ 2a(2a-c) } }=2a(2a−b)(2a−a)(2a−c)×2a(2a−c)(2a−a)(2a−b)

               = 2

Therefore, Cot\frac { B }{ 2 } Cot\frac { C }{ 2 } = 2Cot2BCot2C=2

Answer 2

The value of cot[b/2]cot[c/2] is 10/9.

Given:

A triangle is given by 19a=b+c.

To Find:

The value of cot[b/2]cot[c/2].

Solution:

From the trigonometric properties, $cot\frac{b}{2}= \sqrt{\frac{s(s-b)}{(s-a)(s-c)} }$ and

$cot\frac{c}{2}= \sqrt{\frac{s(s-c)}{(s-a)(s-b)} }$

Multiply the above-written formulas.

$cot\frac{b}{2}\times cot\frac{c}{2}= \sqrt{\frac{s(s-b)}{(s-a)(s-c)} }\times \sqrt{\frac{s(s-c)}{(s-a)(s-b)} }\\=\frac{s}{(s-a)}\,\,\,...(1)$

Also, $s=\frac{a+b+c}{2}$.

From 19a=b+c, substitute the value of 'b+c' into $s=\frac{a+b+c}{2}$ and simplify.

$s=\frac{a+19a}{2}\\ =\frac{20a}{2}\\ =10a$

Now substitute s=10a into equation (1) and simplify.

$cot\frac{b}{2}\times cot\frac{c}{2}=\frac{10a}{(10a-a)}\\=\frac{10a}{9a}\\ =\frac{10}{9}$

Thus, the value of cot[b/2]cot[c/2] is 10/9.

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