Question

in ∆ABC,angle B=90°.if AB =BC = 2cm and AC= 2√2 cm, find the value of sec C, cosec C, and cot C​

Answer 1

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \put(2, 2){\line(0, 1){2}} \put(2,4){\line( 1, - 1){2}} \put(2, 2){\line(1,0){2}} \put(2,1.8){\sf B} \put(2,4){\sf A} \put(4,1.8){\sf C} \put(3,1.8){\sf 2cm} \put(3,3){ \sf 2 \sqrt{2}cm } \put(1.1,3){\sf 2cm} \put(2,2.1){\sf 90^{ \circ} } \end{picture}$

Given:

• B = 90°
• AB = BC = 2cm
• AC = 2√2cm

Find:

• Sec C, Cosec C and Cot C

Solution:

we, know that

$\sf \dashrightarrow \sin C = \dfrac{H}{B}$

$\sf \dashrightarrow \sin C = \dfrac{2 \sqrt{2} }{2}$

$\sf \dashrightarrow \sin C = \sqrt{2}$

$\therefore$ sin C = √2

Now,

$\sf \dashrightarrow \cosec C = \dfrac{H}{P}$

$\sf \dashrightarrow \cosec C = \dfrac{2 \sqrt{2} }{2}$

$\sf \dashrightarrow \cosec C = \sqrt{2}$

$\therefore$ cosec C = √2

Now,

$\sf \dashrightarrow \cot C = \dfrac{B}{P}$

$\sf \dashrightarrow \cot C = \dfrac{2}{2}$

$\sf \dashrightarrow \cot C = 1$

$\therefore$ cot C = 1

_________________________

$\red{\sf \to \sin C = \sqrt{2}}$

$\green{ \sf \to \cosec C = \sqrt{2}}$

$\pink{\sf \to \cot C = 1}$