# Question #90e4b

Dec 15, 2016

$\frac{\sqrt{21}}{2}$

#### Explanation:

$T a n B = \text{opposite / adjecent}$

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

Tan B = $\text{AC / CB}$

find AC using Pythagoras theorem which is
$A {C}^{2} = {5}^{2} - {2}^{2}$
AC = $\sqrt{{5}^{2} - {2}^{2}}$
AC = $\sqrt{21}$

Put the values of AC and CB in the equation

Tan B = $\frac{\sqrt{21}}{2}$

Dec 15, 2016

$\tan B = \frac{\sqrt{21}}{2}$ (answer $f$)

#### Explanation:

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

$A B$ = hypotenuse (H)
$C B$ = adjacent (A)
$A C$ = opposite (O)

$\tan B$ = O/A = $A C$/$C B$

find CB:

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

$C {B}^{2}$=$A {B}^{2}$-$A {C}^{2}$

$= {5}^{2} - {2}^{2}$

$= 25 - 4$

$= 21$

$C B = \sqrt{21}$

$\tan B$ = $A C$/$C B$ = $\frac{\sqrt{21}}{2}$

$\tan B = \frac{\sqrt{21}}{2}$