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.

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Answer 1

Given:-

In ∆ABC ,

• AC = 17 units
• BC = 8 units
• right angled at $\angle$B

To Find:-

all the trigonometric ratios of angle C.

Solution:-

Applying Pythagoras Theorem in ∆ABC

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

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

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

Putting Values -

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

$: \implies$$AB = \sqrt{ 289 - 64}$

$: \implies$$AB = \sqrt{ 225}$

$: \implies$$AB = 15 units$

Now Finding ratios -

$\sin C= \frac{P}{H} = \frac{AB}{AC} = \boxed{\frac{15}{17}}$

$\cos C = \frac{B}{H} = \frac{BC}{AC} = \boxed{\frac{8}{17}}$

$\tan C = \frac{P}{B} = \frac{AB}{BC} = \boxed{\frac{15}{8}}$

$\cosec C= \frac{H}{P} = \frac{AC}{AB} = \boxed{\frac{17}{15} }$

$\sec C= \frac{H}{B} = \frac{AC}{BC} = \boxed{\frac{17}{8} }$

$\cot C = \frac{B}{P} = \frac{BC}{AB} = \boxed{\frac{8}{15}}$

Where,

• P = Perpendicular
• B = Base
• H = hypotenuse

and we are done :)