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Trigonometric Ratios

In Δ ABC ...

Question

In $Δ$ $A B C$ the sides opposite to angles $A, B, C$ are denoted by $a, b, c$ respectively.
If $H$ is the orthocentre of triangle ABC, then $A H$ is equal to?

2 R cos A

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2 R sin A

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a cot A

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\frac{2 a b c}{Δ} cos A

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Solution

The correct options are
A $2 R cos A$
C $a cot A$
$\frac{1}{2} (A D) a = \frac{1}{2} b c sin A$

$B H = \frac{C cos B}{sin (90 - C)}$

$\frac{A H}{sin (90 - A)} = \frac{B H}{sin (90 - B)}$

$\frac{A H}{cos A} = \frac{C cos B}{cos C \times cos B}$

$A H = \frac{C cos A}{cos C}$

$A H = a cot A (∵ \frac{a}{sin A} = \frac{c}{sin C})$

$A H = 2 R cos A (\frac{a}{sin A} = 2 R)$

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