We can find x using the law of cosines in the triangle AOB. We need the cosBAO. First we draw the line segments AO=radius AC=√40 and the perpendicular OK to the chord AC. From the orthogonal triangle AOK we calculate cosKAO ,sinKAO From the orthogonal triangle ABC we calculate cosCAB , sinCAB Then we take the type of trigonometry cosBAO=cos(BAC-KAO)=cosBACcosKAO+sinBACsinKAO=7√50/50. FInally from the law of cosines x=√26

When we find angle BCO = 45º, we can use cosine rule in triangle BCO : x² = BC² + OC² - 2.BC.OC.cos45º x² = 36+50 - 2(6)(sqr.(50))(0.5sqr(2)) x² = 86 - 6(sqr(100)) x² = 86 - 60 = 26 x = sqr(26). Thank you for your sharing

Drop a perpendicular from O to M on BC Set MO=a, MB=b In triangle OMB a^2 + b^2 = x^2 [1] In triangle OMC a^2 + (6-b)^2 = 50 [2] In a right triangle on hypotenuse OA (a+2)^2 + b^2 = 50 [3] Subtract [2] from [3] 4a + 4 + 12b - 36 = 0 4a + 12b = 32 a + 3b = 8 a = 8 - 3b Substitute for a in [2] (8-3b)^2 + (6-b)^2 = 50 9b^2 - 48b + 64 + b^2 - 12b + 36 = 50 10b^2 - 60b + 50 = 0 b^2 - 6b + 5 = 0 (b-1)(b-5) = 0 If b=5, a=8-15 = -7, not allowed So b=1, a=5 x^2 = 1 + 25 x = sqrt(26)

If O is the center of the circle and sq. root of 50 is the radius. The segment BC cannot be equal to 6. Because sq rt of 50 = 7.07. Otherwise the drawing is misrepresented.

Great stuff - amazing works, wonderful!!❤

Da carnot abbiamo 50=4+x^2-4xcos(90+OBC) e 50=36+x^2-12xcos(OBC)...da cui risulta x=√74,x=√26(corretta)

Draw a circle with line segment AC as its diameter (which also passes through point B). You can solve it more easily.

Si trazamos op perpendicular a bc, op es igual a raiz de 50 menos 2, angulo en c 46 grados y X= 5.20

Scale of pic is gonna be way off √50 > 6. It's gonna make 1 look more than twice as long as 5.

❤🎉

Why AD = DC

WTF? how can the radius be shorter than BC? If BC is 6, the radius is 7.07

I could not find a quick and easy solution, but this works, and gives me an excuse to use my favourite formula… Extend C0 to P on the circle In triangle ABC, AC^2 = SQRT 40, by Pythagoras in triangle ABC CP IS diameter, so CAP = 90 ACB = atn(2/6) = BAP (right-angle triangles) CosBAP = cos BCA = 3/sqrt10 CP^2 = (2*CO)^2 = 4*50 = 200 AP^2 = 200 - 40 = 160 In triangle APB, by cosine rule 3/sqrt10 = (4+160-y^2)/4*sqrt160 BP = Sqrt116 OP = OC = radius = SQRT50 In triangles BPO and CBO, using the cos (supplement)) rule (see alternative method below) Cos BOP = - Cos COB (x^2 + 50 -BP^2)/2*x*50 = (6^2 - 50 -x^2)/2*x*50 2x^2 = 36 - 100 + 116 x^2 = 26 Alternative method: In triangle BAP, by sin rule 2/sinBPA = BP/sinBAP Angle APB = 3.366 degrees In 90 triangle APC APC = atn(sqrt40/sqrt160) = 26.565 degrees BPO = APC - APB = 23.2 degrees In triangle BOP, by cosine rule Cos BPO = (y^2 +50 -x^2)/2*sqrt50*BP x^2 = 26