Is This That Tricky That Most People Just Give up On It!???

[00:00:00]Now let's draw a line from the center of the semicircle to meet point C on the circumference of the semicircle and also another line from the center or to meet point D at the circumference of the circle just like this. So as you realize OC so this is the center O. So OC is the radius also O D is the radius which whereby also O B is the radius. So this line is equals to this line is also equals to this line which is equals to the radius. Now in triangle OBD since OB is equals to OD that implies that OBD is an isocles triangle and therefore if this angle here is 57° that implies that this angle will also be 57° and remember interior angles of a triangle add up to 180°. Therefore this angle D O B will be equals to 180°us 57° - 57° which is equals to 180° - 114° which is equals to 66°.

[00:01:30]So this is 66°.

[00:01:35]Now in triangle B, triangle B O D and triangle C O D. This triangle here and this triangle here B O is equals to C O. So this line is equals to this line which are both the radius. also OD is equals to O D which they share and BD is equals to DC.

[00:02:15]So this implies that triangle B O D is congruent to triangle C O D.

[00:02:30]Therefore implying that angle B O D is equals to angle C O D.

[00:02:40]So this angle here is also 66° also.

[00:02:46]So yeah it's 66°.

[00:02:51]Now we know that in any circle, we know that in any given circle where we have a chord AB and this is the center center O and we have this angle that is subended at the circumference from the code.

[00:03:19]this if this angle is theta then this angle here at the circumference will be theta / by two so that if in reference to that in our diagram here so given the code B if this is the code BC now this angle from the center is So angle B O C is 66° + 66° which is equals to 132°.

[00:03:59]So this is 3132°.

[00:04:01]Now from the same code CB there is an angle at the subended at the circumference. So from this code so from the code B up to this circum point of circumference A.

[00:04:16]So this implies that angle X from this it will be equals to this angle which is 132 / 2 which will be equals to 66° and therefore X is equ= to 66°.

[00:04:39]Let's look at another way in which you could have found the value of X.

[00:04:45]So let's start by joining point A to D just like this. We know that in any circle if this is the diameter and any angle that is subended from the circumference from the diameter to the circumference of the circle is 90°. Even if we had it like this, this angle would be 90°.

[00:05:14]That implies that since this is the diameter of this semicircle and the angle is being subended at the circumference from the diameter, that implies that this angle here is 90°.

[00:05:30]Now in this triangle AD angle B A D this angle here is equals to 180°us 90° - 57° which is equals to 180 - 90 is 90° - 57° which is equals to 33°.

[00:06:03]Therefore, this angle is 33°.

[00:06:08]Now we know that in any circle if you have a code say A B and you have another code C D and then we have this code to this and also this code to this.

[00:06:36]Now if this code is equals this this is equals to this. If this angle here is theta that implies that this angle will also be theta. If this A code A is equals to code C D then this angle will be equals to this angle.

[00:07:06]Now since in our figure BD is equals to C D that implies that angle B A D is equals to angle C A D.

[00:07:23]So this angle is equals to this angle.

[00:07:26]So if this is 33° this will also be 33° and therefore x will be equals to this angle plus this angle which is 33° + 33° and therefore x is equals to 66°.

[00:07:48]So that's how you solve. Thank you so much for watching.