Question

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

Answer 1

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`

Answer 2

Please see below.

Explanation:

.

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`