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.
@@RAG981Both are correct. As ∠OBC = 65°, and as ∆COB and ∆DOC are congruent isosceles triangles, then ∠BCO = ∠OCD = ∠CDO = 65°, and ∠COB = ∠DOC = 180°-2(65°) = 50°. That means ∠DOB = 2(50°) = 100° and ∠AOD = 180°-100° = 80°, and you can get that ∠DAB = x = 50° either by recognizing that ∆AOD is isosceles and thus ∠ODA = ∠DAO = (180°-80°)/2 = 50°, or that ∠DAB = ∠DOB/2 = 100°/2 = 50°. Either one works.
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
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°.
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
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.
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!✨ 😂😂😂
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
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° ]
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.
1/Connect OC, OD. The triangles DOC and COB are conguent so the angles DOC and COB are conguent too. --> x = 1/2 angle DOB= angle COB Focus on the isosceles triangle COB -> x= 180 - ( 2. 65 ) =50 degrees😅😅😅
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
Let's find x: . .. ... .... ..... The triangle OBC is an isosceles triangle. So we can conclude: ∠OCB = ∠OBC = 65° ∠BOC = 180° − ∠OCB − ∠OBC = 180° − 65° − 65° = 50° The triangles OBC and OCD are congruent (OB=OC=OD and BC=CD). So we obtain: ∠OCD = ∠ODC = 65° ∠COD = 50° The triangle OAD is an isosceles triangle as well. So we can calculate the value of x as follows: ∠AOD = 180° − ∠BOC − ∠COD = 180° − 50° − 50° = 80° x = ∠OAD = ∠ADO = (180° − ∠AOD)/2 = (180° − 80°)/2 = 100°/2 = 50° Best regards from Germany
Solution: If OB is the radius and OC, as well, is the radius, then OBC is an isosceles triangle. Therefore, the angle OCB is 65°. Now, let's trace the radius OD and since BC is equal to CD, then the triangles OBC and OCD are congruents. Thus, the angle C is 65° + 65° = 130° Finaly, the quadrilateral ABCD is a Cyclic Quadrilateral and the opposite angles ad up 180° Therefore: A + C = 180° x + 130° = 180° x = 50° ✅
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.
Actually x is half of the 100, angle at centre is twice angle at circumference.
Ce qui est plus facile
ഞാനും ഈ രീതിയിലാണ് ചെയ്തത്
I like it.
@@RAG981Both are correct. As ∠OBC = 65°, and as ∆COB and ∆DOC are congruent isosceles triangles, then ∠BCO = ∠OCD = ∠CDO = 65°, and ∠COB = ∠DOC = 180°-2(65°) = 50°. That means ∠DOB = 2(50°) = 100° and ∠AOD = 180°-100° = 80°, and you can get that ∠DAB = x = 50° either by recognizing that ∆AOD is isosceles and thus ∠ODA = ∠DAO = (180°-80°)/2 = 50°, or that ∠DAB = ∠DOB/2 = 100°/2 = 50°. Either one works.
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
This is the simplest method.
@zhaoyue5953 Thanks.
OC is parallel to AD
x = 180°-2*65°= 50° ( Solved √ )
x = ½α = ½.2(180°-2*65°) = ½.100°
x = 50° (Solved √)
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
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°.
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
Thank you!
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.
Yes… that is what I figured out.
Simpler solution : AOD angle is 130 degrees (center angle doubles of 65 ) in the isosceles triangle of AOD x=(180-130)/2=25
You got it wrong bro...
OBC = 65
Not OBD
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!✨ 😂😂😂
I did the same solution by inspection as well.
I too think it's an easier approach
X + X + 65 + 65 + 65 + 65 = 360
X = 50
🎉
m
Good to be here particularly for me,no doubt.
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
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° ]
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.
Bom dia Professor
Obrigado pelo exercício
Grato
180°ABCDO/90=2ABCDO (ABCDO ➖ 2ABCDO+2).
Could you please tell me the name of software/platform on which you make the videos?
1/Connect OC, OD.
The triangles DOC and COB are conguent so the angles DOC and COB are conguent too.
--> x = 1/2 angle DOB= angle COB
Focus on the isosceles triangle COB
-> x= 180 - ( 2. 65 ) =50 degrees😅😅😅
Do u teach tuition
STEP-BY-STEP RESOLUTION PROPOSAL :
01) Observe that : OA = OD = OC = OB = R
02) Isosceles Triangle [OBC] = Isoceles Triangle [OCD]
03) This two Isosceles Triangles are : (65º ; 65º ; 50º)
04) Triangle [OAD] is Isosceles with Angles (Xº ; Xº ; Yº)
05) Angle (BOD) = 50º + 50º = 100º. So, Angle (AOD) = (180º - 100º) = 80º
06) Yº = 80º
07) As Triangle [AOD] is Isosceles we have :
08) 2Xº + 80º = 180º ; 2Xº = 180º - 80º ; 2Xº = 100º ; Xº = 100º / 2 ; Xº = 50º
Therefore,
OUR BEST ANSWER :
Angle X = 50º
NOTE:
As Central Angle (BOD) = 100º then Inscribed Angle (BAD) = 100º / 2 = 50º
This was my approach as well. No over-complicated esoteric theorems, just some Isosceles Triangles.
Excellent!👍
Thanks for sharing ❤️
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
X = [360° - 4(65°)] ÷ 2 = 100° ÷ 2 = 50°
X = 2 × (90° - 65°) = 50°
Excellent!
Thanks for sharing ❤️
My way of solution ▶
Let's consider the triangle ΔOAB, this is an isosceles triangle:
[OA]=[BO]= r
[AB]= a
∠OAB= 65°
∠ABO= 65°
⇒
∠BOA= 180°-2*65°
∠BOA= 50°
By applying the sinus theorem, we get:
sin(50°)/a= sin(65°)/r= sin(65°)/r
a= r*sin(50°)/sin(65°)
stept-2) Let's consider the isosceles triangle ΔOBC:
∠COB= Θ
[OB]= [CO]= r
[BC]= [AB]
[BC]= a
[BC]= r*sin(50°)/sin(65°)
∠OBC= y
∠BCO= y
∠COB= Θ
By applying the cosine theorem we get:
a²= r²+r²-2r²*cos(Θ)
[r*sin(50°)/sin(65°)]²= 2r² - 2r²cos(Θ)
r²*(0,845236523)² = 2r²(1-cos(Θ))
0,71442478*r²= 2r²(1-cos(Θ))
1-cos(Θ)= 0,35721239
cos(Θ)= 0,642787609
⇒
Θ= 50°
step-3) Let's consider the isosceles triangle ΔOCD :
[DO]=[OC]= r
∠ CDO= ∠OCD = x
∠ DOC= α
α= 180° - Θ - ∠BOA
Θ= 50°
∠BOA= 50°
⇒
α= 80°
α+2x= 180°
80°+2x= 180°
⇒
x= 50° ✅
Let's find x:
.
..
...
....
.....
The triangle OBC is an isosceles triangle. So we can conclude:
∠OCB = ∠OBC = 65°
∠BOC = 180° − ∠OCB − ∠OBC = 180° − 65° − 65° = 50°
The triangles OBC and OCD are congruent (OB=OC=OD and BC=CD). So we obtain:
∠OCD = ∠ODC = 65°
∠COD = 50°
The triangle OAD is an isosceles triangle as well. So we can calculate the value of x as follows:
∠AOD = 180° − ∠BOC − ∠COD = 180° − 50° − 50° = 80°
x = ∠OAD = ∠ADO = (180° − ∠AOD)/2 = (180° − 80°)/2 = 100°/2 = 50°
Best regards from Germany
Simply 180-2×65=50.😊
No
You got it wrong
65 is ABC not ABD
？@@gurujeetparihar？？？？？？
X=50°.❤
🙂
😀
Thanks ❤️
My apologies I just realize why it was necessary because apart of the 65• was cut off! Thanks beautiful math problem
You are welcome! Glad you enjoyed it. 😊
Thanks for the feedback ❤️
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Solution:
If OB is the radius and OC, as well, is the radius, then OBC is an isosceles triangle.
Therefore, the angle OCB is 65°.
Now, let's trace the radius OD and since BC is equal to CD, then the triangles OBC and OCD are congruents. Thus, the angle C is 65° + 65° = 130°
Finaly, the quadrilateral ABCD is a Cyclic Quadrilateral and the opposite angles ad up 180°
Therefore:
A + C = 180°
x + 130° = 180°
x = 50° ✅
50 derece . 25 sn lik soru
Thanks for the feedback ❤️
OB =OC =>OBC=BCO=65°=>BOC=180°-65°×2=50°
∆OBC=∆OCD=>COD=BOC=50°
BOD=50°+50°=100°=>AOD=180°-100°=80°
AO=OD=>X=ADO=>X=(180°-80°)/2=50°
That's all
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