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
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°.❤❤❤
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!✨
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!!!
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. ...🙂
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°
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
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°
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°
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°