An "Hourglassoid"

Geometry Level 1

Let $ABCD$ be a square, and $P$ a point inside the square such that $\triangle PCD$ is an equilateral triangle, as shown below. Find the angle $\alpha$ in degrees.

The answer is 15.

3 solutions

Peter van der Linden
Feb 5, 2017

Angle PDC = 60(equilateral triangle), so $ADP = 30 ^\circ,$ . Since AD = DP this makes angle DAP = DPA = 150/2 = 75 degrees. So this leaves $\alpha = 90^\circ - 75^\circ = 15^\circ$ .

Right, once we've identified that DAP is an isosceles triangle with $\angle DAP = \angle APD$ , we can find $\alpha = 90^\circ - \angle DAP$ .

Pi Han Goh - 4 years, 4 months ago

Sir can u explain that how AD=DP ???

Irfan Khan - 4 years, 1 month ago

Sure AD = DC since ABCD is a square. DC = DP since DCP is equilateral.

Staffan Bonnier - 4 years, 1 month ago

Lorenzo Calogero
Apr 25, 2017

The squadre side measures $a$ . The height of PDC is $\sqrt{3}a/2$ , so the height of APB is $a-\sqrt{3}a/2$ . $\alpha = \arctan{[a(1-\frac{\sqrt{3}}{2})/(a/2)]} = \arctan{(2-\sqrt{3})} = \boxed{15^\circ}$

15 degrees is right and not just the figure 15....this is a poorly intelligent site .

Jainanan Ram - 4 years, 1 month ago

Betty BellaItalia
Apr 26, 2017

Correct answer is: 15°

Let AB=a. Because △PCD is an equilateral triangle, ⇒ |DP|=|PC|=|DC|=a and ∡ADP=90°-60°=30°. So, in the triangle △APD⇒|AD|=|DP|=a and ∡PAD=∡APD=(180°-30°)/2=75°. Then, the angle ∡α=90°-∡PAD=90°-75°=15°.

looks familiar in gillesanias reviewer

Ramiel To-ong - 4 years ago

An "Hourglassoid"

The answer is 15.

3 solutions

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