Another easy trigo

Geometry Level 3

$\angle{ABD} + \angle{ACD} = 90^\circ$

Then, AD = ?

\sqrt{ pq}

\frac{p}{q}

\sqrt{ \frac{p}{q} }

2 solutions

Sakanksha Deo
Mar 10, 2015

Let $\angle{ABD} = x$ $and \angle{ACD} = y$

Now,

$\angle{ABD} + \angle{ACD} = 90^\circ$

$\Rightarrow y = 90 - x$

Now,

$tanx = \frac{ AD }{ p }$ ....... (1)

$tany = tan(90 - x) = cotx = \frac{ AD }{ q }$ ....... (2)

Multiplying (1) and (2),

$1 = \frac{ (AD)^{2} }{ pq }$

$\Rightarrow AD = \sqrt{ pq }$

I was too lazy to do all that math, so I decided instead to eliminate the answers that could not be correct, based on their units.

Since $p$ and $q$ are distances (lengths), I will consider them to be measured in meters. The answer we are looking for should also be a distance, so its unit should be meters.

B: $\frac{p}{q}$ The units (meters) cancel out, so this is a proportion (a ratio).

C: $p q$ The units here are $m^2$ so this in an area.

D: $\sqrt{\frac{p}{q}}$ The square-root of a proportion? I don't know what that is, but it's certainly not a distance!

So none of B, C or D can be the correct answer, because none of them produce a value with the unit meters.

A: $\sqrt{p q}$ The square-root of an area is a distance. So this is the only answer which can be valid!

If you had set one of the multiple-choice answers to also have the unit meters. (For example E: $2p - q$ ) then I could not have used this technique, and would have been forced to work harder!

Paul Clark - 6 years, 3 months ago

Its a level 2 question....you don't need to work that hard...just use your Brian and I think that's what you have done

Sakanksha Deo - 6 years, 3 months ago

Michael Fuller
Mar 10, 2015

Another easy trigo

2 solutions

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