Each rectangle is 6cm long and 3cm wide, they share a common diagonal of PQ. How do you show that #tanalpha = 3/4#?

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2 Answers
Apr 23, 2018

I get #tan alpha = tan(pi/2 - 2 arctan(3/6)) = 3/4#

Explanation:

Fun. I can think of a few different ways to see this one. For the horizontal rectangle let's call the top left S and the bottom right R. Let's call the apex of the figure, a corner of the other rectangle, T.

We have congruent angles QPR and QPT.

# tan QPR = tan QPT = frac{ text{opposite} }{ text{adjacent} } = 3/6 = 1/2 #

The tangent double angle formula gives us #tan RPT#

#tan(2x) = frac{ 2 tan x }{1 - tan^2 x}#

#tan RPT = frac{2 ( 1/2) }{1 - (1/2)^2 } = 4/3 #

Now #alpha# is the complementary angle of RPT (they add up to #90^circ#), so

# tan alpha = cot RPT = 3/4 #

Apr 23, 2018

Please see below.

Explanation:

.

enter image source here

Triangles #DeltaABP# and #DeltaCBQ# are right angle triangles that have:

#AP=CQ=3# and

#/_ABP=/_CBQ# because they are vertical angles.

Therefore, the two triangles are congruent.

This means:

#PB=BQ#

Let #AB=x# and #BQ=y# then:

#PB=y#

We know that:

#x+y=6# cm #color(red)(Equation-1)#

In triangle #DeltaABP#:

#y^2=x^2+9# #color(red)(Equation-2)#

Let's solve for #y# from #color(red)(Equation-1)#:

#y=6-x#

Let's plug this into #color(red)(Equation-2)#:

#(6-x)^2=x^2+9#

#36-12x+x^2=x^2+9#

#36-12x=9#

#12x=27#

#x=9/4#

#tanalpha=(AB)/(AP)=x/3=(9/4)/3=9/12=3/4#