In a triangle ABC right angled at B - sin B - A - C -cosec C-tan A - is equal toA- 0B- 1C- 12D- 3

The correct option is B 1 nbsp; ...

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Calculating Heights and Distances

In a triangle...

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In a triangle ABC right angled at B, sin B [(sec A .cot C)/(cosec C.tan A)] is equal to

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0.0

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tan B = \sqrt{3}

sin A + sin B

tan B

=

2 sin B

A +

sin B

A + B + C = π

∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n (A + B + C) & s i n B & c o s C - s i n B & 0 & t a n A c o s (A + B) & - t n a A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

A B C

B

\frac{sec A . sin C - tan A . tan C}{sin B}

$In triangle ABC, internal angle bisector of ∠ A makes an angle θ with side BC . Value of sin θ is equal to$

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