In a triangle ABC with usual notations, A/2+ B/2+ C/2=

Correct Answer is : s2Δ{s²}{Δ}Δs2, where ΔΔΔ is the area of the triangle ABCA B CABC.

{s²}{Δ}, where Δ is the area of the triangle A B C.

s/Δ, where Δ is the area of the triangle ABC .

{Δ}{{s}}, where Δ is the area of the triangle ABC .

Δ, where Δ is the area of the triangle ABC . Using half angle formula, {aligned} A/2 & ={(s-b)(s-c)/s(s-a)} \\& ={{s(s-a)(s-b)(s-c)}{s²(s-a)²}}{aligned} ∴ A/2={Δ}{{s}({s}-a)}, where Δ is area of triangle ABC Similarly, {aligned}& B/2=Δ/s(s-b), C/2=Δ/s(s-c) \\& A/2+ B/2+ C/2 \\& ={1}{ A/2}+{1}{ B/2}+{1}{ C/2} \\& =s(s-a)/Δ+s(s-b)/Δ+s(s-c)/Δ \\& =s/Δ[3 s-(a+b+c)] \\& =s/Δ(3 s-2 s)={s²}{Δ}{aligned}

In a triangle ABC with usual notations, A/2+ B/2+ C/2=

Correct Answer is : s2Δ{s²}{Δ}Δs2, where ΔΔΔ is the area of the triangle ABCA B CABC.

Test Index

MHTCET 23 Apr 2025 Shift 1 Paper

Question : 119 of 150

Marks: +1, -0

In a triangle ABC with usual notations,

\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=

$\frac{s^{2}}{\Delta}$ , where $\Delta$ is the area of the triangle $A B C$ .
$\frac{s}{\Delta}$ , where $\Delta$ is the area of the triangle ABC .
$\frac{\Delta}{\mathrm{s}}$ , where $\Delta$ is the area of the triangle ABC .
$\Delta$ , where $\Delta$ is the area of the triangle ABC .
Using half angle formula,
$\begin{aligned}\tan \frac{A}{2} & =\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \\& =\sqrt{\frac{s(s-a)(s-b)(s-c)}{s^{2}(s-a)^{2}}}\end{aligned}$
$\therefore \quad \tan \frac{A}{2}=\frac{\Delta}{\mathrm{s}(\mathrm{s}-a)}$ , where $\Delta$ is area of triangle ABC
Similarly,
$\begin{aligned}& \tan \frac{B}{2}=\frac{\Delta}{s(s-b)}, \tan \frac{C}{2}=\frac{\Delta}{s(s-c)} \\& \cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2} \\& =\frac{1}{\tan \frac{A}{2}}+\frac{1}{\tan \frac{B}{2}}+\frac{1}{\tan \frac{C}{2}} \\& =\frac{s(s-a)}{\Delta}+\frac{s(s-b)}{\Delta}+\frac{s(s-c)}{\Delta} \\& =\frac{s}{\Delta}[3 s-(a+b+c)] \\& =\frac{s}{\Delta}(3 s-2 s)=\frac{s^{2}}{\Delta}\end{aligned}$

Validate

Solution:

Concept:

Use half-angle formula

\tan \dfrac{A}{2} = \dfrac{\Delta}{s(s-a)}

and similarly for

B

and

C

Explanation:

Given

\Delta

is the area and

s

is the semiperimeter, we have:

\tan\frac{A}{2}=\frac{\Delta}{s(s-a)},\quad \tan\frac{B}{2}=\frac{\Delta}{s(s-b)},\quad \tan\frac{C}{2}=\frac{\Delta}{s(s-c)}

Therefore,

\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2} = \frac{s(s-a)}{\Delta}+\frac{s(s-b)}{\Delta}+\frac{s(s-c)}{\Delta}

= \frac{s}{\Delta}[3s-(a+b+c)]

Since

a+b+c=2s

= \frac{s}{\Delta}(3s-2s) = \frac{s^2}{\Delta}

Answer:

\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2} = \frac{s^2}{\Delta}

Correct option: A.

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