Can you find the value of angle X? | (Semicircle) |  

Very good Thanks Sir Thanks PreMath With my respects ❤❤❤❤

Угол x, вписанный в окружность и равен половине градусной меры дуги CB=(180-37°*2-37°*2)/2=32°. X=16°.

ΔACD is isosceles with AD = CD. Therefore, the angles opposite those sides are equal, so

Sum of the external angeles of a semicircle is 90 degress, so arc AB =90 degrees . AD=37 and DC=37 (isoceles triangle) so arc CB=16 3rd year junior high school problem in Japan

Sum of opposite angles of circumscribed quadrilateral, is 180° (x + 37°) + (37°+90°) = 180° x = 16° ( Solved √ )

DAC = DCA = 37 (due to isosceles triangle base angles being the same) BCD = BCA + DCA = 90 (due to angles in semicircle 90 degrees) + 37 = 127 DAB = DAC + CAB = 37+x BCD + DAB = 180 (opposite angles in cyclic quadrilateral) 127+37+x = 180 x = 16°

DACº=37º ; ACBº=90º → CABº=180º-90º-37º-37º=16º =X. Gracias y saludos.

AD=CD So DCA=DAC=37° ADC+DCA+DAC=180° ADC+37°+37°=180° So ADC=180°-74°=106° Connect B to C So ACB=90° ADC+CBA=180° So CBA=180°-106°=74° So x=90°-74°=16°.❤❤❤

Equally, the opposite angles of a cyclic quadrilateral add up to 180 degrees. Thus X = 180 - 37 - 37 - 90 = 16.

angle en A = angle en C ,angle AoD =angle de la corde AD au centre= 2 angle ACD=74°, angle COD= angle de la corde DC au centre =2 angle CAD = 74°,triangle AOC isocèle angle AOC = 2x 74°=148°, x= (180°-148°)/2=16° et AC est perpenediculaire à OD.

I did it differently. Arc ADCB is a semicircle of 180deg. Forgot the name of the theorem, but the angle DCA is 37deg, so that cuts a 74deg arc AD. Same for angle DAC being 37deg and cuts a 74deg arc DC. So arc ADC is 74 + 74 = 148, leaving arc CB as 180 - 148 = 32deg. Same dealy now, that angle CAB = 16deg. ✨Magic!✨

Т.к.

The answr is x = 16°. Because of thise two theorems, Thales Theorem and the theorem of cyclic quadrilaterals, this was a problem easier than it looks. Also I think that there should be a playlist devoted to problems that make use of those two insights which btw make attempts at other deductions feel superfluous. That is how good this video is!!!

Very satisfying solution. Thank you.

16 180 - 37 + 37 = 106 angle D 180 - 106 = 74 angle B as angle D+ B = 180 (Cyclic quadilateral) Draw a line from C to B, angle P=90 (Thales Theorem) Angle C + P = 37 + 90 = 127 Angle A = 37 (given) Angle A + x= 37 + x Angle C+ P + A+ x = 180 (Cyclic quadilatre) 127 + 37 + x = 180 164 + x = 180 x = 180 - 164 x =16 Answer

I read the Description before solving and had to put my color blind glasses on and proceeded too teach myself how to find the area of a (Yellow shaded triangle?) so @ 1:54 when I started thinking outside the box I never recovered from losing sight of a yellow triangle. Only thing that came to mind was Thales Theorem. Then proceeded to solve for x. ...🙂

Bom dia Mestre Obrigado por mais uma aula Forte Abraço do Rio de Janeiro

Triangle ABC is right at C, so

Draw radii OD and OC. Let the point where OD intersects AC be E. As CD = DA, as OA = OC = r, and as OD is common, triangles ∆AOD and ∆DOC are congruent by side-side-side congruency. As CD = DA and as OA = OC, ∆CDA and ∆AOC are isosceles triangles. As ∠CDO = ∠ODA and ∠AOD = ∠DOC, OD bisects both ∆AOC and ∆CDA, and as AC is a chord, then OD must be a perpendicular bisector of AC and all four angles at intersection point E are 90°. As angle ∠ACD is on the circumference and subtends arc AD, then angle ∠AOD, which also subtends arc AD and is at the center of the semicircle, must equal twice ∠ACD, or 2(37°) = 74°. In triangle ∆OEA, as ∠AOE = 74°, and as ∠OEA = 90° then ∠EAO = x = 180°-90°-74° = 16°. x = 16°

Thank you!

Connect BD and use Thales theorem to show

Let's find x: . .. ... .... ..... The triangle ACD is an isosceles triangle (AD=CD). So we can conclude: ∠CAD = ∠ACD = 37° According to the theorem of Thales the triangle ABC is a right triangle (∠ACB=90°). By combining this triangle and the isosceles triangle ACD we obtain the quadrilateral ABCD. Since this quadrilateral has a circumscribed circle, we know that the sum of two opposite angles is 180°: ∠BAD + ∠BCD = 180° (∠BAC + ∠CAD) + (∠ACB + ∠ACD) = 180° (x + 37°) + (90° + 37°) = 180° ⇒ x = 180° − 90° − 2*37° = 16° Best regards from Germany

Because AO = DO, 37+x must be half the angle ADC, i.e. 37+x = 53

My way of solution ▶ In this isosceles triangle ΔACD [DA]= [DC] ⇒ ∠DAC= ∠ACD ∠DAC= 37° ∠ACD= 37° Segment AD, s(AD) s(AD)= 2*∠ACD s(AD)= 2*37° s(AD)= 74° s(DC)= 2*∠DAC s(DC)= 2*37° s(DC)= 74° s(AB)= 180° s(CB)= 180°- s(AD)- s(DC) s(CB)= 180° - 74° - 74° s(CB)= 32° ∠CAD= s(CB)/2 ∠CAD= 32°/2 ∠CAD= 16° ⇒ x= 16°

AngleADC = 106° (easy). Angle AOC = 180° -2.x in triangle AOC. We draw E as O is the middle of [D,E], E is on the circle and angle AEC is half of angle AOC, so it is 90° -x. Now angle ADC + angle AEC = 180°, so 106° + (90° - x) = 180°, and that gives that x = 16°

Radius OD intersects Chord AC Perpendicularly at K. Since Chord AD subtends 74° at O, in the Right triangle AKO, X= 90-74=16°

φ = 30° → sin⁡(3φ) = 1; ∆ ACD → DAC = ACD = 37φ/30 → CDA = 53φ/15 → AOC = 106φ/15 → 12φ - AOC = 74φ/15 = COA → CAO = OCA = (6φ - 74φ/15)/2 = x = 8φ/15 or: CDO = ODA = 53φ/30 → ∆ AOD → DAO = ODA = 53φ/30 → x = 53φ/30 - 37φ/30 = 8φ/15

Solution: Since ADC is an isosceles triangle, the angle CAD worths 37°, as well, and the angle ADC worths 106° Therefore, the central angle is twice the segment angle. In this way, the central angle runs through an angle of 212° (106° x 2 = 212°). Connecting Point "O" to Point "C", we create a new isosceles triangle OAC, forming the angle COB = 32° (212° - 180°). Thus, the angle AOC is 148° (180° - 32°). The sum of the internal angles of a triangle is 180°, thus, x + x + 148° = 180°, therefore x = 16°

16. So. Easy

I have, at least, two ways of solving this Problem. First Way : 01) Close the Semicircle to get a Circle. 02) AD = CD 03) Triangle [ACD] Is an Isosceles Traingle. So: 04) Angle (ACD) = Angle (CAD) = 37º 05) Angle (ADC) = 180º - (37º * 2) ; Angle (ADC) = 180º - 74º ; Angle (ADC) = 106º 06) Angle (CAD) is an inscribed Angle. 07) Angle (DOC) is an Central Angle, of 37º * 2 = 74º 08) Central Angle (DOA) = Central Angle (DOC) = 74º 09) 74º + 74º = 148º 10) Central Angle (CAB) = 180º - 148º ; Central Angle (CAB) = 32º 11) So, Inscribed Angle (CAB) = 32º / 2 ; Inscribed Angle (CAB) = X = 16º Second Way : 01) As we can easily see, the Inscribed Angle (ADC) = 106º 02) So the Central Angle (AOC) = 212º 03) As Central Angle (AOB) = 180º 04) 212º - 180º = 32º 05) Inscribed Angle (CAB) = X = 16º Thus, The Answer is : Angle X = 16º

DAOC is a kite, so its diagonals AC and OD are perpendicular. Being angle in D = 106° => ADO = 106/2 = 53 triangle ADO is isosceles (being OA and OD radii) so ADO = DAO = 53° finally X = 53° - 37° = 16°

I added Arcs AD & DC (74+74=148), subtracted from 180 to get Arc CB =32, therefore angle X= 16.

106+(90-x)=180..x=16

Let me tell you a easy and short method angle AOD =2ACD, AO =OD ,DAO =53, X=16

74×2+2x=180, 74+x=90, x=90-74=16. Actually in general x=90-2×37.😅

X=16

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53 ^0 ?

x=16