In ABC- right angled at B - - ACB -60- Values of C and cosec C are -A- Sec C-C- Cosec C - A -N-AB- Sec C - AL - BC - Cosec C - A - AC- Sec C - AC - BCC - Cosec C - m - B CD- Sec C-A C-A B- Cosec C-A -B C

The correct option is B Sec C = AC/BC ,Cosec C = AC/AB Sec C = 1/cosC =Hypotenuse/Base = AC/BC and cosec C = 1/sinC = Hypotenuse/Perpendicular AC/AB ...

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Standard IX

Mathematics

Opposite & Adjacent Sides in a Right Angled Triangle

In ABC, right...

Question

In $△$ ABC, right angled at B, $∠$ ACB = $60^{\circ}$ , Values of sec C and cosec C are :

Sec C = $\frac{A C}{B C}$ ,Cosec C = $\frac{A C}{A B}$

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Sec C = $\frac{A C}{B C}$ ,Cosec C = $\frac{A C}{B C}$

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Sec C = $\frac{A C}{A B}$ , Cosec C = $\frac{A C}{A B}$

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Sec C = $\frac{A C}{A B}$ ,Cosec C = $\frac{A C}{B C}$

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Solution

The correct option is A
Sec C = $\frac{A C}{B C}$ ,Cosec C = $\frac{A C}{A B}$

Sec C = $\frac{1}{c o s C}$ = $\frac{H y p o t e n u s e}{B a s e}$

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

and cosec C = $\frac{1}{s i n C}$ = $\frac{H y p o t e n u s e}{P e r p e n d i c u l a r}$

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

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In △ ABC, right angled at B, ∠ACB = 60∘, Values of sec C and cosec C are :

The correct option is A Sec C = ACBC ,Cosec C = ACAB Sec C = 1cosC =HypotenuseBase = ACBC and cosec C = 1sinC = HypotenusePerpendicular ACAB

In $△$ ABC, right angled at B, $∠$ ACB = $60^{\circ}$ , Values of sec C and cosec C are :

The correct option is A
Sec C = $\frac{A C}{B C}$ ,Cosec C = $\frac{A C}{A B}$

Sec C = $\frac{1}{c o s C}$ = $\frac{H y p o t e n u s e}{B a s e}$

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

and cosec C = $\frac{1}{s i n C}$ = $\frac{H y p o t e n u s e}{P e r p e n d i c u l a r}$

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