Triangles inside a square
is a square and
is a point inside it so that
and angle
is 15 degrees. What is the magnitude of angle
in degrees?
Bonus: Try solving without the use of trigonometry.
The answer is 60.
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One way is to draw a line parallel to BC through P and solve using trigonometry. But to do it geometrically, the method of angle chasing doesn't seem to work. Hence we make use of inequalities:
First, call ang. APB as 'x' and deduce the other angles in terms of x as shown in the diagram. It is easily seen that triangle APB is isosceles with AP = BP
Now let's assume AP > AB, and noting that an angle opposite to a longer side in a triangle is larger than an angle opposite to a shorter side, we can say
ang. ABP > ang. APB
90 - x/2 > x
x/2 < 30
Reversing the signs and adding 105 to both sides we get:
105 - x/2 > 105 - 30
105 - x/2 > 75
ang. APD > ang. ADP which must mean that AD > AP
Since we began with the assumption that AP > AB, this must mean AD > AP > AB which is impossible because AD and AB are sides of a square and hence must be equal.
Making the exact opposite assumption as the one we made earlier i.e. assuming AP < AB, we can show in exactly the same way that it leads to AD < AP < AB which is again impossible because AD = AB.
As AP > AB and AP < AB are both impossible, then the only other option is AP = AB which makes the triangle APB equilateral.
Thus, ang. APB = x = 60 degrees