Isosceles triangle inside square | Find the angle | Advanced math problems | Mathematics

This problem is an application of the properties of congruent triangles. This problem introduces a different technique for solving the problems to find angles from a given figure.

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#maths #mathematics #square #isoscelestriangle #triangle #angle #advancedmathproblems #calculation #problems #geometry

[davidseed]

very interesting. i difnt see that approach.
Note ADP is an equilateral triangle .
So tan15⁰ = 2-sqrt3

[jimlocke]

For this solution method, I make use of the known ratios of sides for a 15° - 75°- 90° triangle and for a 30° - 60° - 90° triangle. These ratios have been derived and published in many places. This solution method does not make use of Problem Analysis's very clever construction of the second isosceles triangle ΔQDC, congruent to ΔPCD.

Drop a perpendicular from point P to BC. Label the intersection as point M. Drop a perpendicular from point P to CD. Label the point of intersection as point N. Note that <BPM = 75° by sum of angles of a triangle (ΔBPM) = 180°. By same reasoning with ΔCPM, <CPM = 75°. Length BP = length CP by virtue of ΔBPC being isosceles. Therefore, ΔBPM and ΔCPM are congruent by side - angle - side (second side common) and length MB = length MC. Let the length of the side of the square be designated s. Length BC is the length of the side of the square, so length MC is half the length of the side of the square, or s/2.

<PCN = 75° because it is 90°- <PCM. <CPN = 15° by sum of angles of a triangle, ΔCPN, = 180°. Therefore, ΔPCN and ΔCPM are congruent by angle - side - angle, the side PC being common. Length PN, therefore, equals length MC = s/2.

The ratio of short side to long side of a 15° - 75°- 90° triangle is (2 - √3):1. So, length CN = (2 - √3)s/2 = s - s(√3)/2. Length DN = length DC - length CN = s - (s - s(√3)/2) = s(√3)/2). The ratio of sides PN to ND of ΔPND is (s/2)/( s(√3)/2)) = 1/(√3) The known ratio of short side to long side for a 30° - 60° - 90° triangle is 1:(√3). So, ΔPND is a 30° - 60° - 90° triangle and <PDN = 30°. <ADP, the ? angle, and <PDN must sum of 90°, so the ? angle is 60°.

[firstnamelastname]

if circle with center A and radius AB and circle with center D and radius DC cross in Q it is rather easy to see that angles QBC and QCB are 15 degrees. So P is Q.

[shadrana]

At 2:16,
Let ABCD be a unit square,
Draw FE parallel to CD through P,
Join A and P,
AP=PD mirror image about
tan15=(2-sqrt3) standard
Consider the blue triangle BPC,
In the bottom half, is triangle PEC,
Angle ECP=15 deg, EP=1/2 and PC
tan15=(2-sqrt3)=EP/EC,
(2-sqrt3)=2EP
EP=(2-sqrt3)/2
Draw a vertical PP' to intersect DC,
You now have a right triangle PP'D



DP=AP=AD=1 from (1) and (2)
Triangle APD is equilateral with angles DAP, APD and ADP=60deg.
Hence angle ADP=60 deg. and that's our answer.

[pi]

If side of square is two
Tanx =(2-Tan15)/1=1.73
X=tan inv(1.73)=60°

[-Keith-]

This solution doesn't work when triangle PQC isn't an equilateral triangle, so it's a bit of a cheating trick to solve it this way because you'd need to know ahead of time that a 15 degree isosceles triangle along the side of a square would result in equilateral triangle PQC. So it's a quick solution, but only under this one condition.

If you try it again with 37 degree matching angles for your isosceles triangle, you end up with angle BPC = 106 degrees, angle PCQ = 16 degrees, and angle PQC = QPC = 82 degrees. Then your remaining triangle DPQ has 3 different side lengths, and nothing to relate them to the other triangles inside the square.

The more reliable real-world way to approach the solution for any triangle angle other than 15 degrees would be simple trigonometry, which will solve the problem without having to change methods when this trick doesn't work.

Assign an arbitrary value of 1 to the side length of the square, since the angular measurements will be the same for any size of square. Draw a line from P to the right middle side of the square. We'll call this point R. Triangle CPR has 1 known angle and 1 known side, so we can use tan(15) = PR/0.5, simplified to line length PR = tan(15) * 0.5 = approximately 0.13397. Save the result into your calculator memory without rounding or dropping any decimals.

Now draw a line straight down from point P to the bottom edge of the square, and label this point S. This creates lines DS and SC. Line SC = line PR. Line length DS = 1 - SC = approximately 0.866025 (again, save the number into your calculator memory without rounding or dropping decimals). This gives us a right angle triangle DSP, with two known sides: PS = 0.5, and DS which is already mentioned. Tan(PDS) = 0.5/DS = 0.57735, therefore angle PDS = 30 degrees. Therefore angle ADP = 60 degrees.