In a triangle ABC,right angled at B,if Ab=4cm and BC=3cm,find all six trigonometric ratios of angle A
Answers
Answer:
Appropriate Question :-
In triangle ABC, is right angled at B. If AC = 5cm, BC = 3cm and AB = 4cm. Find all the six trigonometric identities related to angle A.
Answer :-
★ FIGURE
Refer the attachment.
Firstly, we know, There are six trigonometric identities :-
\sf{sine (sin \: A ) = \dfrac{Perpendicular}{Hypotenuse} \implies \dfrac{BC}{AC}}sine(sinA)=
Hypotenuse
Perpendicular
⟹
AC
BC
\sf{cosine (cos \: A) = \dfrac{Base}{Hypotenuse} \implies \dfrac{AB}{AC}}cosine(cosA)=
Hypotenuse
Base
⟹
AC
AB
\sf{tangent (tan \: A) = \dfrac{Perpendicular}{Base} \implies \dfrac{BC}{AB}}tangent(tanA)=
Base
Perpendicular
⟹
AB
BC
\sf{cotangent (cot \: A) = \dfrac{Base}{Perpendicular} \implies \dfrac{AB}{BC}}cotangent(cotA)=
Perpendicular
Base
⟹
BC
AB
\sf{secant (sec \: A) = \dfrac{Hypotenuse}{Base} \implies \dfrac{AC}{AB}}secant(secA)=
Base
Hypotenuse
⟹
AB
AC
\sf{cosecant (cosec \: A ) = \dfrac{Hypotenuse}{Perpendicular} \implies \dfrac{AC}{BC}}cosecant(cosecA)=
Perpendicular
Hypotenuse
⟹
BC
AC
We have, Given
AC = 5cm
BC = 3cm
AB = 4cm
\begin{gathered} \\ \end{gathered}
∴ According the question,
\begin{gathered} \\ \\ \end{gathered}
sine ( sin A )
\sf{ \longrightarrow \: sine (sin \: A ) = \dfrac{BC}{AC}}⟶sine(sinA)=
AC
BC
\sf{ \longrightarrow \: \dfrac{BC}{AC}}⟶
AC
BC
\sf{ \longrightarrow \: \dfrac{3}{5} \: \: \: \: \bigstar}⟶
5
3
★
\begin{gathered} \\ \end{gathered}
cosine ( cos A )
\sf{ \longrightarrow \: cosine (cos \: A ) = \dfrac{AB}{AC}}⟶cosine(cosA)=
AC
AB
\sf{ \longrightarrow \: \dfrac{AB}{AC}}⟶
AC
AB
\sf{ \longrightarrow \: \dfrac{4}{5} \: \: \: \: \bigstar}⟶
5
4
★
\begin{gathered} \\ \end{gathered}
tangent ( tan A )
\sf{ \longrightarrow \: tangent (tan \: A) = \dfrac{BC}{AB}}⟶tangent(tanA)=
AB
BC
\sf{ \longrightarrow \: \dfrac{BC}{AB}}⟶
AB
BC
\sf{ \longrightarrow \: \dfrac{3}{4} \: \: \: \: \bigstar}⟶
4
3
★
\begin{gathered} \\ \end{gathered}
cotangent ( cot A )
\sf{ \longrightarrow \: cotangent (cot \: A) = \dfrac{AB}{BC}}⟶cotangent(cotA)=
BC
AB
\sf{ \longrightarrow \: \dfrac{AB}{BC}}⟶
BC
AB
\sf{ \longrightarrow \: \dfrac{4}{3} \: \: \: \: \bigstar}⟶
3
4
★
\begin{gathered} \\ \end{gathered}
secant ( sec A )
\sf{ \longrightarrow \: secant (sec \: A) = \dfrac{AC}{AB}}⟶secant(secA)=
AB
AC
\sf{ \longrightarrow \: \dfrac{AC}{AB}}⟶
AB
AC
\sf{ \longrightarrow \: \dfrac{5}{4} \: \: \: \: \bigstar}⟶
4
5
★
\begin{gathered} \\ \end{gathered}
cosecant ( cosec A )
\sf{ \longrightarrow \: cosecant (cosec \: A ) = \dfrac{AC}{BC}}⟶cosecant(cosecA)=
BC
AC
\sf{ \longrightarrow \: \dfrac{AC}{BC}}⟶
BC
AC
\sf{ \longrightarrow \: \dfrac{5}{3} \: \: \: \: \bigstar}⟶
3
5
★