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Properties of Triangle question

2021 · Shift 1 · Q37

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JEE Advanced
Mathematics
Properties of Triangle
2021 · Shift 1 · Q37

Properties of Triangle question

2021 · Shift 1 · Q37

JEE AdvancedMathematicsProperties of TriangleNumerical+4 / −1

In a triangle ABC, let AB =

\sqrt {23}

, BC = 3 and CA = 4. Then the value of

{{\cot A + \cot C} \over {\cot B}}

is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 2

Given sides

In $\triangle ABC$ :

$AB = \sqrt{23}$
$BC = 3$
$CA = 4$

Using standard notation:

side opposite $A$ is $a = BC = 3$
side opposite $B$ is $b = CA = 4$
side opposite $C$ is $c = AB = \sqrt{23}$

We need to find

\frac{\cot A + \cot C}{\cot B}.

Use area formula for cotangents

For any triangle,

\Delta = \frac12 bc\sin A

so

\sin A = \frac{2\Delta}{bc}.

Also,

\cos A = \frac{b^2 + c^2 - a^2}{2bc}.

Hence,

\cot A = \frac{\cos A}{\sin A} = \frac{\frac{b^2+c^2-a^2}{2bc}}{\frac{2\Delta}{bc}} = \frac{b^2+c^2-a^2}{4\Delta}.

Similarly,

\cot B = \frac{c^2+a^2-b^2}{4\Delta}, \qquad \cot C = \frac{a^2+b^2-c^2}{4\Delta}.

Compute each cotangent numerator

Given

a^2=9,\quad b^2=16,\quad c^2=23.

So,

\cot A = \frac{16+23-9}{4\Delta} = \frac{30}{4\Delta},

\cot B = \frac{23+9-16}{4\Delta} = \frac{16}{4\Delta},

\cot C = \frac{9+16-23}{4\Delta} = \frac{2}{4\Delta}.

Form the required expression

\frac{\cot A + \cot C}{\cot B} = \frac{\frac{30}{4\Delta} + \frac{2}{4\Delta}}{\frac{16}{4\Delta}} = \frac{32}{16} = 2.

Final answer

\boxed{2}

Comparison with stored answer

Stored correct answer: $2$ .

Our derived answer matches the stored answer.

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