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 ...

You visited us 8 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

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

Suggest Corrections

Similar questions

In $Δ$ ABC, right angled at B, expressions for sec C and cosec C is

Q. In a

△

ABC, right angled at B, AC - AB = 1 unit, and BC = 7 units.
What is

cosec C

$△$ ABC, right angled at B, $∠$ ACB = $60^{\circ}$ , expressions for sec 60 and cosec 60 is

Q. If
A,B,C are the interior angles of triangle ABC, then prove that

cosec (\frac{C + A}{2}) = sec (\frac{B}{2})

Q. In a right angled triangle ABC right angled at B,

5 \times s i n A = 3

.
Find the value of cosC + tanA + cosecC.

Join BYJU'S Learning Program

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

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