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Can you find the angle X? | (Semicircle) | #math #maths | #geometry

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[thomasgreene]

A simpler solution is to recognize triangle BOC is isosceles, so the angle BOC is 50 deg. The triangles BOC and COD are congruent, so the angle BOD is 100 deg. This makes the angle AOD 80 deg. The triangle AOD is isosceles, so the angle x is 50 deg.

[jimlocke]

<CBA = 65° is an inscribed angle, so it subtends an arc with double the angle, or 130°. That arc is AC. Arc AB is 180°. Arc BC is AB - AC = 180° - 130° = 50°. Equal length chords subtend equal arcs, so CD = BC. Arc BD = BC + CD = 50° + 50° = 100°. x = <DAB is an inscribed angle subtending arc BD, so is half the arc's measure, or 50°. So, x = 50°, as PreMath also found.

[PrithwirajSen-njqq]

Join AC. Angle ACB=90 degrees being inscribed in semi circle.
Angle CAB =180-65-90 =25 degrees
Angle CAB = angle CAD= 25 degrees
( inscribed angles are on equal chord)
Hence angle x = 25+25=50 degrees

[ОльгаСоломашенко-ьы]

Соединяем точки D и C с центром окружности. Получаем два равных равнобедренных треугольника с улами при вершине O равными 180°-65°*2=50°.< 0:57 DOB=50°2=100°. Это центральный угол окружности, угол x вписанный в окружности и равен половине этого центрального угла x=50°.

[مطبخالشاطرسراج]

A simple solution: draw AC, m(arc DB) = 2x, since angle DAB is inscribed and equals half the subtended arc, but m(arc BC) = ½ m(arc DB) = x because DC = BC, and therefore AC bisects angle A, and m(ACB) equals 90 degrees, so we have an equation: x/2 + 65 + 90 = 180, solving it will give us x = 50

[alexundre]

Bom dia Professor
Obrigado pelo exercício
Grato

[allanflippin]

The key is that we have a cyclical quadrilateral. I found a very simple way with math you can solve in your head. Add lines to form triangles OCB and OCD. It's easily shown that these are isosceles triangles. Therefore angles OCB and OCD are both 65 and the total angle C is 130. x = 180 - 130 so x = 50.

[burgazadaeczanesi]

Simpler solution : AOD angle is 130 degrees (center angle doubles of 65 ) in the isosceles triangle of AOD x=(180-130)/2=25

[PrithwirajSen-njqq]

Join OC.and OD. Now triangle OBC is isosceles.
Angle BOC = 180- 2*65 = 50 degs
Triangles OCD and OBC r congruent.
Then ang COD =50 degs
Ang BOD = 50+50=100 degrees
Inscribed Angle x = 1/2 of centre angle BOD =50 degrees

[bkp_s]

Good to be here particularly for me, no doubt.

[marioalb]

OC is parallel to AD
x = 180°-2*65°= 50° ( Solved √ )

[PrithwirajSen-njqq]

Another method
Join OC and OD
. Now triangle OBC is isosceles. Ang OBC = ang OCB =65 degs
Triangles OBC and OCD r congruent. Hence
ang OCD =65 degrees
Ang BCD = 65*2 = 130 degrees
Then x = 180 -130=50 degrees.

[marioalb]

x = ½α = ½.2(180°-2*65°) = ½.100°
x = 50° (Solved √)

[davidstecchi]

m<ADC + m<ABC = 180, m<ABC= 65 therefore m<ADC =115, m<ADB = 90 so m<BDC = 25

[engralsaffar]

i connected OD and OC, to get 2 equilateral triangles, with 65 degrees for angles ODC, OCD and OCB, this makes angle DOB =100 degrees (360-4*65)
angle x is the angle at the circumference and is therefore half of the angle at the center DOB, therefore x=50 degrees

[joeschmo]

It took me like 30sec from just the thumbnail. Draw 2 radii to make triangles COB and DOC, as well as AOD. They''re all isosceles with 2 radii as sides. DOC has 65deg and another 65deg angle, making 130, from 180 leaves a 50deg angle. Same for triangle COB (CB and DC are equal). So you got 2 50deg angles making angle DOB 100deg. That leaves angle DOA as a supplementary angle of 80deg. 180 - 80 = 100deg for 2 angles x and x (again, isosceles). Split the difference, and that makes each x=50deg.
Oh, and ✨Magic!✨ 😂😂😂

[RealQinnMalloryu]

180°ABCDO/90=2ABCDO (ABCDO ➖ 2ABCDO+2).

[quigonkenny]

As points A, B, C, and D all appear on the circumference of semicircle O, ABCD is a cyclic quadrilateral and its opposite angles must sum to 180°. That means rhat ∠ABC and ∠CDA sum to 180°, and ∠BCD and ∠DAB sum to 180°. ∠ABC is given as 65°, so ∠CDA = 180°-65° = 115°.

Draw BD. As A and B are points at the ends of a diameter and D is a point on the circumference, then by Thales' Theorem, ∠BDA = 90°. As ∠CDA = 115°, ∠CDB = 115°-90° = 25°.

As BC = CD, ∆BCD is an isosceles triangle, so ∠DBC = ∠CDB = 25°. As the internal angles of a triangle must sum to 180°, that means that ∠BCD = 180°-2(25°) = 130°. Also, as ∠ABC = 65° and ∠DBC = 25°, ∠ABD = 65°-25° = 40°.

We now have two methods to determine the value of ∠DAB:

As ∠DAB and ∠BCD must sum to 180° and ∠BCD = 130°, x = 180°-130° = 50°.

To confirm, as the the internal angles of a triangle must sum to 180°, ∠ABD = 40°, and ∠BDA = 90°, then ∠DAB = x = 180°-(90°+40) = 50°.

[ x = 50° ]

[vikramsinghshekhawat]

Could you please tell me the name of software/platform on which you make the videos?

[PrithwirajSen-njqq]

Another method

A) ABCD is a cyclic quadrilateral.
Hence angle C= 180 -x
Join DO.
AOD is an isosceles triangle


B) Ang BAD=Ang ADO =x

Join BD.
BCD is an isosceles triangle
C) ang BD C= angle CDB (1)

BOD is an isosceles triangle.
D) ang ODB = ang OBD (2)

Adding 1 and 2 we get E) E) angle
ODC = ang OBC =65 degs

F) Again Angle DOB =2x
Now angles
DOB +OBC +BCD +ODC =360
>2x +65 +180 -x +65=360
> x =50
x = 50 degrees