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.


Please answer my question​

Answer 1

$\huge\rm { ☆_!! Question !_! ☆}$

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

$\huge\rm { ☆_!! Solution !_! ☆}$

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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Answer 2

Remember:-

• sin(∅) = (Opposite Side)/(Hypotenuse)
• cos(∅) = (Adjacent Side)/(Hypotenuse)
• tan(∅) = (Opposite Side)/(Adjacent Side)
• cosec(∅) = 1/sin
• sec(∅) = 1/cos
• cot(∅) = 1/tan

Answer:-

Finding the length of AC:- (Hypotenuse)

We know,

(Hypotenuse)² = (Base)² + (Height)²

[Pythagoras Theorem]

Thus,

AC² = BC² + AB²

→ AB² = AC² - BC²

→ AB² = (AC + BC)(AC - BC)

→ AB² = (17 + 8)(17 - 8)

→ AB² = √[(25)(9)]

→ AB² = 5 • 3

→ AB² = 15 units

Trigonometric Ratios:- (of C)

Adjacent Side = BC

Hypotenuse = AC

Opposite side = AB

• sin C = AB/AC = 15/17
• cos C = BC/AC = 8/17
• tan C = AB/BC = 15/8
• cosec C = AC/AB = 17/15
• sec C = AC/BC = 17/8
• cot C = BC/AB = 8/15

Note: Refer to the diagram above by @JiangXue.