I didn’t use trigonometry because if triangle AEC is isosceles the base angles are equal therefore angle CAE is x+15. So x+15 +x+15 + 90 = 180 ie 2x + 30 = 90 so 2x = 60 and therefore x=30.
I bring CK perpendicular to AB. Hence AD=DB=k, AB=2k, and CK=h. From the right triangle CKB because B=30 then CB=2h. (1)._ From the right triangle CKD because angle D=45 then CD=h (root 2). (2)._ I bring AH perpendicular to BC. From the right triangle AHB because angle B=30 then AH=k. (3)._ In the triangle ACB, CD is a median, and from the median theorem I find AC=k (root 2). (4),_ the right triangle AHC {because (3) (4)}, is isosceles AH=CH and (angle CAH)=(angle ACH) =45 and because angle DCB=15 {ACB-DCB=ACD} tending X=(angle ACD)=30
30 Let DB =1 (any other number could be used), then AD also = 1 Now you have a Side 1, an Angle 135, and another angle 30 from triangle BDC Using the Law of Sine a/b = sine A / sine B, the unknown side b = 1.932 For triangle ACD, in which X = angle C, we have a Side of 1, another side of 1.932, and a 45-degree angle . Using the law of cosine, yield 30 degrees for X So first let AD = any length, and you you have ASA, then the new side will lead to SAS.
@@hussainfawzer From 3'50, the triangle ADE is equilateral because AD=DE and angleADE=60° therefore on the one hand AD=ED=EA. On the other hand the triangle CDE is isosceles at E because angleDCE=angleCDE=15° and therefore EC=ED. We conclude that ED=EA=EC and therefore E is the circumcenter of the triangle ADC.
Because the first method is by geometry method. And the second method is by trigonometry (sine rule). In the video, it is taking time because I am writing every step. If you solve it by yourself, you will get that sine rule and m-n cot theorem both will take nearly equal time.
Miarę tego kąta można łatwo policzyć z twierdzenia sinusów i twierdzenia cosinusów Z twierdzenia sinusów w BCD można wyrazić długości boków trójkąta BCD za pomocą długości boku BD Z twierdzenia cosinusów w ACD wyrażamy długość boku AC za pomocą długości boku BD a następnie także z twierdzenia cosinusów wyznaczamy cosinus kąta x Zadanie dość łatwe
I didn’t use trigonometry because if triangle AEC is isosceles the base angles are equal therefore angle CAE is x+15. So x+15 +x+15 + 90 = 180 ie 2x + 30 = 90 so 2x = 60 and therefore x=30.
I bring CK perpendicular to AB. Hence AD=DB=k, AB=2k, and CK=h. From the right triangle CKB because B=30 then CB=2h. (1)._ From the right triangle CKD because angle D=45 then CD=h (root 2). (2)._ I bring AH perpendicular to BC. From the right triangle AHB because angle B=30 then AH=k. (3)._ In the triangle ACB, CD is a median, and from the median theorem I find AC=k (root 2). (4),_ the right triangle AHC {because (3) (4)}, is isosceles AH=CH and (angle CAH)=(angle ACH) =45 and because angle DCB=15 {ACB-DCB=ACD} tending X=(angle ACD)=30
Although the second method is complicated but it is very interesting.
Thank you.
30
Let DB =1 (any other number could be used), then AD also = 1
Now you have a Side 1, an Angle 135, and another angle 30 from triangle BDC
Using the Law of Sine a/b = sine A / sine B, the unknown side b = 1.932
For triangle ACD, in which X = angle C, we have
a Side of 1, another side of 1.932, and a 45-degree angle . Using the law of cosine,
yield 30 degrees for X
So first let AD = any length, and you you have ASA, then the new side will
lead to SAS.
Triangle 🔺️ ½=a/b(sin)= degree 30, Angle ACD(CDB)
In the first method it is easy to prove that E is the circumcenter of triangle ACD and therefore AngleACD=1/2*AngleAED=60°/2=30°
How to prove point E is the circumcenter of the triangle ACD ?
@@hussainfawzer From 3'50, the triangle ADE is equilateral because AD=DE and angleADE=60° therefore on the one hand AD=ED=EA. On the other hand the triangle CDE is isosceles at E because angleDCE=angleCDE=15° and therefore EC=ED. We conclude that ED=EA=EC and therefore E is the circumcenter of the triangle ADC.
Excellent, as always.
Thanks 🙂
Why don't you ever use m-n cot theorem?
It's much easier that way
Because the first method is by geometry method.
And the second method is by trigonometry (sine rule). In the video, it is taking time because I am writing every step. If you solve it by yourself, you will get that sine rule and m-n cot theorem both will take nearly equal time.
@@MathBooster still m n cot theorem is faster
if AD = BD, then isn't ACD = DCB?
obfuscating the question and never overwriting on the figure
v good explain😍🥰
Thank you 🙂
1
Życzę sukcesów w osu i w matmie
@@mikolaj79 haha
Miarę tego kąta można łatwo policzyć z twierdzenia sinusów i twierdzenia cosinusów
Z twierdzenia sinusów w BCD można wyrazić długości boków trójkąta BCD za pomocą długości boku BD
Z twierdzenia cosinusów w ACD wyrażamy długość boku AC za pomocą długości boku BD
a następnie także z twierdzenia cosinusów wyznaczamy cosinus kąta x
Zadanie dość łatwe
3🥰🥰
30 degree
sir😍🥰
math🥰🥰🥰
Are you Indian ?
31 degree