Question

IF SIN A=3/4, CALCULATE COSA AND TAN A.

Answer 1

SinA = 3/4
Perpendicular = 3k/4k
~~~~~~~~~~~~
Hypotenuse

By Pythagoras Theorem ,
(H)^2 = (B)^2 +(P)^2
(4k)^2 = (B)^2 + (3k)^2
16k2 - 9k2 = B^2
Root7 k = B


CosA = B/H
CosA = Root7 k/4k
CosA = Root7 /4


TanA = P/B
TanA = 3k/Root7 k
TanA = 3/Root7

Answer 2

$Given, sinA= </p><p>4</p><p>3$

$⇒ </p><p>DC</p><p>BC$

$= </p><p>4</p><p>3$

⇒BC=3k and AC=4k

where k is the constant of proportionality.

By Pythagoras theorem, we have

$AB </p><p>2$$=AC </p><p>2$ $−BC </p><p>2$ $=(4k) </p><p>2$ $−(3k) </p><p>2$

$=7k </p><p>2$

$⇒AB= </p><p>7$

k

So, cosA=

$AC/</p><p>AB$

$= </p><p>4</p><p>7$

$k</p><p> </p><p> = </p><p>√7/4$

And tanA=

$AB/</p><p>BC$

=

7 $</p><p>3$

$= </p><p>3/√7$

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