Question #90e4b

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Answer 1 · 2016-12-15T18:36:23

$sqrt21 / 2$

$Tan B = "opposite / adjecent"$

As our angle under consideration is angle B. We will take the sides adjacent to B and opposite to B.

Tan B = $"AC / CB"$

find AC using Pythagoras theorem which is
$AC^2 = 5^2 - 2^2$
AC = $sqrt(5^2 - 2^2)$
AC = $sqrt21$

Put the values of AC and CB in the equation

Tan B = $sqrt21 / 2$

Answer 2 · 2016-12-15T18:39:18

$tan B = sqrt(21)/2$ (answer $f$ )

first, label the sides of the triangle in relation to angle B:

$AB$ = hypotenuse (H)
$CB$ = adjacent (A)
$AC$ = opposite (O)

$tan B$ = O/A = $AC$ / $CB$

find CB:

$a^2+b^2=c^2$

$CB^2$ = $AB^2$ - $AC^2$

$= 5^2 - 2^2$

$=25-4$

$=21$

$CB = sqrt(21)$

$tan B$ = $AC$ / $CB$ = $sqrt(21)/2$

$tan B = sqrt(21)/2$