Find Angle X in this Semicircle | Step-by-Step Tutorial

[procash]

Superb Sir

Almost solved it myself mentally, except for a small error in calculation

I egarly wait for your "exciting" videos. For Maths lovers, these videos are really exciting indeed

[kshitizmangalbajracharya]

If you join BE and use the properties of cyclic quadrilateral, then it will be much easier since angle AEB is 90°.

[spiderjump]

There is a shorter method. In right triangle AEB, angle ABE = 90-65=25.
Therefore angle EBC=
71-25= 46.
X = 180-46 = 134( angles in opp segments are supplementary)

[Escviitash]

I have noticed that many people do not name the angles correctly when posting their solutions.
When naming angles the first letter represents the point/vertex on the starting leg, and the last letter represent the point/vertex on the ending leg, and since angles are measured anticlockwise, interior angles must be named: point/vertex on right leg, origin, point/vertex on the left leg.
E.g the given angle of 71° is the measurement of ∠CBO, not of ∠OBC which is = 289°.
It can be a bit conusing that the leftmost letter represents the point/vertex on the right leg and vs, but that is just how it is.

[armandoguerra]

The sum of the internal angles of a polygon is: 180 ° (N-2) where N = number of sides. In our case 5 sides, so:
sum of internal angles 180 (5-2) = 540 °.
therefore the sum of the angles OED + ODE = 540 ° -71 ° -140 ° -69 ° -65 ° -65 ° = 130 °
whereby ODE = 65 ° and x = 65 ° + 69 ° = 134 °

[denismongeau]

If you consider the 4 triangles shown in the video, the sum of all their angles i 4x180=720. The sum of the angles at the base is 180, so the sum of the remaining angles is 720 - 180 = 540 = 65 + 130 + x + 140 + 71. So x = 540 - 65 - 130 - 140 - 71 = 134 and that is the answer.

[swinkscalibur]

I haven't seen anyone else post this solution, but if you draw in the segment OD, you can use two applications of the Central Angle Theorem. First we use it to find that angle BOD is 80 degrees (360-2(140)=80), then after finding angle AOD is 100 degrees, we use the CAT again to find angle AED is 130 degrees ((360-100)/2=130). Finally we use the Interior Angles of a Polygon Theorem to find the final angle of pentagon ABCDE is 134 degrees.

[sameerqureshi-khcc]

Sir you are an amazing mathematician, love from Pakistan 🌹🌹🌹🌹

[MrDcwithrow]

By inscribed angle theorem angle A implies arc EB is 130, therefore arc AE is 50, and angle B implies arc AC is 142, leaving arc CB as 38. Finally arc EC going around the bottom of the circle is 180+50+38=268, , therefore angle D is half that =134.

[michaeldakin]

The line segments, together with the diameter of the circle, form the pentagonal ABCDE… the sum of internal angles of a polygon is ([number of sides]-2) x 180… it’s good to draw in the radii to find the equal angles of each of the isosceles triangles, but then it’s just a matter of subtracting the total of the known angles from 540 (being the total of the internal angles of the pentagonal), which just leaves a number equal to 2x, so just divide by 2…

[davidfromstow]

Very interesting with another elegant solution

[simonharris]

You don't need to calculate the angles at the centre, as the total of all internal angles of a 5 sided shape sum to 540. If you subtract the 2 x 71 and the 2 x 69 and the 2 x 65 from that, you have 65 each for the last 2 angles along the circumference.

[krislegends]

X = 134? Connecting the radius to each angle on the circumference will form several isosceles triangles.

[math]

very well explained bro, thanks for sharing

[sidimohamedbenelmalih]

Notice that x don't depends of 140

[eecrin]

the peripheral angle, which sees the diameter, is 90 degrees. Combine A with C with a dash. The ACB angle is 90 degrees. The angle of CAB is 90-71=19 degrees, and in the same way the angle of AEB, which combine E with B, is 90 degrees. The EBA angle is 25 degrees. the angle in the circle is equal to half of the arc it sees. it happens that 19.2=38 and 25.2=50 degrees. The semicircle is 180 degrees. 50+38+180/2 =134 deg

[pedroloures]

I had the same aproach! Beautiful problem!

[billcame]

I got 134, too. I figured out all the angles based on all the inner isosceles triangles (all with two sides of radius). I stopped and restarted again because I was very thrown off by it not being to scale (as was mentioned)...as in it isn't even close (note the 2 equal inner triangles on the left).

[xellosblackforest]

When you drew the radius ...

I realised the solution is easier that I planned

[mustafizrahman]

I have solved it very easily. Your question is very interesting. However, the answer is 134 degrees.