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#