If $△$ ABC is an acute angled triangle and BD is perpendicular to AC, then the value of $\frac{sin A}{sin C}$

\frac{A B}{B C}

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\frac{B C}{A B}

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\frac{B C}{C D}

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\frac{C D}{B C}

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Solution

The correct option is B $\frac{B C}{A B}$
Given In $△ A B C$ $B D ⊥ A B$

From $△ B A D,$

$sin A = \frac{B D}{A B}$

From $△ B D C,$

$sin C = \frac{B D}{B C}$

$\frac{sin A}{sin C} = \frac{\frac{B D}{A B}}{\frac{B D}{B C}} = \frac{B C}{A B}$

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Similar questions

△ A B C

A D

B C

A C^{2} = A B^{2} + B C^{2} - 2 B C \times B D

B C^{2} + C D^{2} = 2 A C \times C D

A B C, A D

B C

A B^{2} - A C^{2} = B D^{2} - C D^{2}

△ A B C

{A B}^{2} + {C D}^{2} = {A C}^{2} + {B D}^{2}

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