Question

In triangle ABC,Right angled at A,if AB=5cm,BC=13,Find sinB, cosB, tanB.​

Answer 1

Step-by-step explanation:

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Answer 2

Given:-

• BC of triangle ABC is 13 cm.
• AB of triangle ABC is 5 cm.

To find:-

• Value of sin B , cos B and tan B.

Solution:-

$\bigstar \sf \: sin \: \theta = \dfrac{Perpendicular}{Hypotenuse}$

$\sf \: sin \: B = \dfrac{AC}{BC}$

• We do not have AC. So, by Pythagoras theorem we will find AC

Pythagoras theorem is,

(Base)² + (Perpendicular)² = (Hypotenuse)²

Hypotenuse = BC = 13 cm

Base = AB = 5 cm

Perpendicular = AC = ?

Put all values in Pythagoras theorem,

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

$\implies \sf {(5)}^{2} + {(AC)}^{2} = {(13)}^{2}$

$\implies \sf 25 + {(AC)}^{2} = 169$

$\implies \sf {(AC)}^{2} = 169 - 25$

$\implies \sf {(AC)}^{2} = 144$

$\implies \sf AC = \sqrt{144}$

$\implies \sf AC = 12$

AC is 12 cm.

Therefore,

$\bigstar \sf \: sin \: \theta = \dfrac{Perpendicular}{Hypotenuse}$

$\sf \: sin \: B = \dfrac{AC}{BC}$

$\large \boxed {\sf \: sin \: B = \dfrac{12}{13}}$

$\bigstar \sf \: cos \: \theta = \dfrac{Base}{Hypotenuse}$

$\sf \: cos \: B = \dfrac{AB}{BC}$

$\large \boxed {\sf \: cos \: B = \dfrac{5}{13}}$

$\bigstar \sf \: tan \: \theta = \dfrac{Perpendicular}{Base}$

$\sf \: tan \: B = \dfrac{AC}{AB}$

$\large \boxed {\sf \: tan \: B = \dfrac{12}{5}}$

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More ratio's :

$\bigstar \sf \: cot \: \theta = \dfrac{Base}{Perpendicular}$

$\bigstar \sf \: sec \: \theta = \dfrac{Hypotenuse}{Base}$

$\bigstar \sf \: cosec \: \theta = \dfrac{Hypotenuse}{Perpendicular}$