Question

$\large{\underline{\color{green}{Question}}}$

Given sin A = \dfrac{12}{37} , find cos A and tan A.

Please give full explanation.

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Answer 1

Question

Give sin A = 12/37, find the cos A and tan A

Given

• sin A = 12/37

Required to Find

cos A and tan A

Solution

let Triangle ...which

angle is A, a hypotenus (h) is 37 given, perpendicular (L) is 12 and base (B).

so,

Sin A = 12/37

A= 18.92°

Now cos A= B/H

COS( 18.92) × 37 = B

Base (B) = 35.00

Now, cos A = 35/37

cos A = 0.945

Now tan A = sinA/ cosA

= 0.324/ 0.945

Tan A = 0.342. or

= 12/35 ans

Answer 2

$\huge\tt{Answer:-}$

Let a triangle in which

• Angle = A
• Hypotenuse = H
• Perpendicular = P
• Base = B

Now, we are given :

$\sf sin A = \dfrac{12}{37}$

$\sf \implies \dfrac{P}{H} = \dfrac{12}{37}$

$\sf \implies P = 12 \: and \: H = 37$

⠀ ⠀⠀ ⠀ ⠀⠀ ⠀

Now, by Pythagorean theorem,

H² = B² + P²

$\sf \implies 37² = B² + 12²$

$\sf \implies 37² - 12² = B²$

$\sf \implies 1369 - 144 = B²$

$\sf \implies 1225 = B²$

$\sf \implies B² = 1225$

$\sf \implies B² = 35²$

$\sf \implies B = 35$

$\sf \therefore cosA = \dfrac{B}{H}=\dfrac{35}{37}$

$\sf tanA = \dfrac{P}{B} = \dfrac{12}{35}$