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MHTCET 16 May 2024 Shift 2 Paper

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Question : 109 of 150

Marks: +1, -0

In a triangle

\mathrm{ABC}, l(\mathrm{AB})=\sqrt{23}

l(\mathrm{BC})=3

l(\mathrm{CA})=4

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

is

1
2
4
8

Solution:

Concept:

Use the area of the triangle to express cotangents in terms of side lengths.

Explanation:

Let

a = l(\mathrm{BC}) = 3

,

b = l(\mathrm{CA}) = 4

, and

c = l(\mathrm{AB}) = \sqrt{23}

.

For any triangle with area

\Delta

, we have:

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

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

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

Therefore,

\cot A + \cot C = \frac{2b^2}{4\Delta} = \frac{b^2}{2\Delta}

.

So,

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

.

Substitute the given side lengths:

\frac{2(4)^2}{(\sqrt{23})^2 + 3^2 - 4^2} = \frac{32}{23+9-16} = \frac{32}{16} = 2

.

Answer:

2

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