Question

Consider right triangle ABC, right angled at B. If AC = 17 units and BC = 8 units determine all the trigonometric ratios of angle C.

Answer 1

Given:-

• AC (Hyp.) ⇒ 17 units

• BC (Base) ⇒ 8 units

Formula Used:-

Pythagoras Theorem:- (Hyp.)² ⇒ (Base)² + (Perp.)²

→ (AC)² ⇒ (BC)² + (AB)²

→ (AC)² - (BC)² ⇒ (AB)²

→ (17)² - (8)² ⇒ (AB)²

→ 289 - 64 ⇒ (AB)²

→ 225 ⇒ (AB)²

AB ⇒ $\sqrt{225}$

AB ⇒ 15 Units ⇒ (Perp.)

• Sin C ⇒ $\frac{Perpendicular}{Hypotenuse}$ ⇒ $\frac{AB}{AC}$ ⇒ $\frac{15}{17}$

• Cos C ⇒ $\frac{Base}{Hypotenuse}$ ⇒ $\frac{BC}{AC}$ ⇒ $\frac{8}{17}$

• Tan C ⇒ $\frac{Perpendicular}{Base}$ ⇒ $\frac{AB}{BC}$ ⇒ $\frac{15}{8}$

• Cot C ⇒ $\frac{Base}{Perpendicular}$ ⇒ $\frac{BC}{AB}$ ⇒ $\frac{8}{15}$

• Sec C ⇒ $\frac{Hypotenuse}{Base}$ ⇒ $\frac{AC}{BC}$ ⇒ $\frac{17}{8}$

• Cosec C ⇒ $\frac{Hypotenuse}{Perpendicular}$ ⇒ $\frac{AC}{AB}$ ⇒ $\frac{17}{15}$

Thank

You

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Answer 2

$\huge\rm { ☆_!! Solution !_! ☆}$

We have-

$\rm { AC ( Hypotenuse) = 17 units }$

$\rm { BC ( Base ) = 8units }$

Applying pythagoras property-

$\rm\blue { {(P)}^{2} + {(B)}^{2} = {(H)}^{2} }$

$\rm\red { {(AB)}^{2} + {(BC)}^{2} = {(AC)}^{2} }$

$\rm { \longrightarrow {(AC)}^{2} - {(BC)}^{2} = {(AB)}^{2} }$

$\rm { \longrightarrow {(AB)}^{2} = {(17)}^{2} - {(8)}^{2} }$

$\rm { \longrightarrow {(AB)}^{2} = 289 - 64}$

$\rm { \longrightarrow {(AB)}^{2} = 225}$

$\rm { \longrightarrow AB = \sqrt{225} = 15units }$

Now , we have -

$\rm { AB ( Perpendicular ) = 15 units }$

$\rm { BC ( Base ) = 8 units }$

$\rm { AC ( Hypotenuse ) = 17 units }$

$\sf { \therefore sin \: C = \dfrac{perpendicular}{hypotenuse}}$

$\sf { \implies sin \: C = \dfrac{AB}{AC} = \dfrac { 15}{17} }$

____

$\sf { \therefore cos \: C = \dfrac{base}{hypotenuse}}$

$\sf { \implies cos \: C = \dfrac{BC}{AC} = \dfrac { 8}{17} }$

____

$\sf { \therefore tan \: C = \dfrac{perpendicular}{base}}$

$\sf { \implies tan \: C = \dfrac{AB}{BC} = \dfrac { 15}{8} }$

____

$\sf { \therefore cot \: C = \dfrac{base}{perpendicular}}$

$\sf { \implies cot \: C = \dfrac{BC}{AB} = \dfrac { 8}{15} }$

____

$\sf { \therefore sec \: C = \dfrac{Hypotenuse}{base}}$

$\sf { \implies sec \: C = \dfrac{AC}{BC} = \dfrac { 17}{8} }$

_____

$\sf { \therefore cosec \: C = \dfrac{Hypotenuse}{perpendicular}}$

$\sf { \implies cosec \: C = \dfrac{AC}{AB} = \dfrac { 17}{15} }$

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Thankyou :)