In the given triangle, The value of SinA is

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

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

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

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sin A =1

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Solution

The correct option is A
sin A = $\frac{B C}{A C}$

sin A = $\frac{P e r p e n d i c u l a r}{H y p o t e n u s e}$ = $\frac{B C}{A C}$

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Similar questions

△ A B C, \frac{sin A}{c} + \frac{sin C}{a} = \frac{a}{b c} + \frac{c}{a b},

△

△ A B C

B

B C = 7 c m, A C - A B = 1

cos A + sin A

P

B C

△ A B C .

(A B + B C + A C) > 2 A P .

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