Question

5. If sec 4A = cosec (A-20°), where 4A is an acute angle, find the value of A.​

Answer 1

Answer:

A = 22°

Step-by-step explanation:

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{sec4A = cosec(A - 20 \degree)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find -   \begin{cases} &\sf{value \: of \: A}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf Formula \: Used -   \begin{cases} &\sf{secx = cosec(90\degree - x)}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\tt \longrightarrow \: sec4A = cosec(A - 20\degree )$

$\tt \longrightarrow \: cosec(90\degree - 4A) = cosec(A - 20\degree)$

$\tt\implies \:90\degree - 4A = A - 20\degree$

$\tt\implies \:A + 4A = 90\degree + 20\degree$

$\tt\implies \:5A = 110\degree$

$\tt\implies \boxed{ \purple{ \bf \: \:A = 22\degree}}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

$\sec(4a) = \cosec(a - 20) \\ \\ \cosec(90 - 4a) = \cosec(a - 20) \\ \\ (90 - 4a) = (a - 20) \\ \\ 90 + 20 = a - 4a \\ \\ 5a = 110 \\ \\ a = \frac{110}{5} \\ \\ a = 22 \: answer \\ \\ \\ {\huge{\boxed{\mathbb{\purple{Stay \: strong}}}}}$