Sir, beautifully & patiently explained. You have an ocean of patience which is the prime requirement of any teacher Even a student weak in Maths will easily understand if u take a class Thanks
In ∆CDB, angle B + 15° = 45° [exterior angle is the sum of interior opposite angles] Therefore angle B = 30° Let AD = DB = a and AC = b In ∆ABC by law of sine 2a/sin(x+15°) = b/sin30° = b/(1/2) = 2b => sin (x+15°) = a/b --------(1) In ∆ ADC a/sin x = b/sin45° = b/(1/√2) = b√2 => √2 sin x = a/b ---------(2) From (1) and (2) √2 sin x = sin (x+15°) √2 sin x = sin x cos 15° + cos x sin 15° (√2-cos 15°) sin x = cos x sin15° sin x / cos x = sin 15° /(√2-cos 15°) tan x = sin 15° / (√2-cos 15°) tan x = 1/√3 ( pl. refer Note below) = tan 30° Thus the unknown angle x = 30° Note: sin 15° = sin (45°-30°) = sin 45° cos 30° - cos 45° sin30° = [1/√2] [ √3/2 - 1/2] = (√3-1)/2√2 Similarly cos 15 = (√3+1)/2√2 √2-cos 15 = √2 - (√3+1)/2√2 = (3-√3)/2√2 = √3(√3-1)/2√2 = √3 sin 15 Hence sin 15° / (√2-cos 15°) = 1/√3
I solved by using trignometry and the law of sines for triangles. With reference to your diagram, let AD=BD= a, and let us designate the length AC=b. Then, for the triangle ACD, a/sinx = b/sin45 ==> sinx = a/(b*sqrt 2). Also for the triangle ACB, 2a/sin(x+15) = b/sin 30 ==> sin(x+15) = a/b = sqrt 2*sinx. Therefore, sinx*cos15+cosx*sin15 = sqrt 2*sinx. If we divide this equation by sinx, it can be reduced to cos15 + cotx*sin15 = sqrt 2 ==> cotx = (sqrt 2 - cos15)/sin15, or tan x = sin15/(sqrt 2 - cos15). The right hand expression can be calculated as 0.57735026919, which happens to be the tan 30 degrees.
True, but the video shows the fundamental or the manual way ehe, even for those who haven't learned trigonometry still can solve the problem... Nice video ☺️
Well, there are many ways of solving this beautiful problem!! 1) using elementary geometry. 2) using sine rule 3) using m-n theorem Using the third method gives u the answer in just 1step...😎😎😉😉 Anyways, thanks for sharing! Love from India!!
M-N theorem is very handy and interesting. Went through its proof. (m+n)cot(theta)=m.cot(alpha)-n.cot(beta). In this problem AD=m DB=n; m=n Theta=135 Alpha = X Beta = 15 Sin(15) = (√3-1)/(2.√2) Cos(15) = (√3+1)/(2.√2) CotX = √3 X = 30. Very nice approach. Today i learnt m-n theorem. Thank you Kusuma I used the second approach
I used the law of sines and cosines. Since line AD = DB it's easier to make an assumption about the length and then we can workout CD using the sine rule and CA using the cosine rule then we can use the sine rule again to find the value of X. But thank you, this was a good exercise.
Exactly.. and using this method, you can find that 2x = 60 in less than 1.5 minutes. That's way more than the allotted time for solving a geometry problem like this in CAT exam. We usually have 25-30 seconds for solving such problems in CAT exam.
Don't know what your age or former occupational title has to do with it. And you "guessed" it correctly? There's no guessing as you couldn't possibly know if you "guess" was correct until someone else DEMOSTRATED it to be correct. Ugh You must have been an excellent engineer. SMH
*Trigonometric solution:* Drop a perpendicular from C to meet the base line at E. Let CE= a and ∠ECA= θ. Since ∠DCE=45°, we have ED=a, and since ∠EBC= 30°, we have EB=a✓3. This gives AD= (✓3-1)a and EA=ED-AD=(2-√3)a. Now Tanθ=EA/CE = 2-√3, which gives us θ=15°. Hence X=30°.
It's over 30 years since I took geometry in university (a Euclidian and non-Euclidian course). I thought it was not possible to find X. I also misinterpreted the double hashes as meaning AD and DB were parallel :D So I didn't get far. I was amazing by the explanation. Such a smart tactic. Thanks for taking the time.
Interesting solution. Another way is drawing CH and proving that A and H are one point, because the other two possible positions of H lead to contradictions with the triangle CHB.
Great puzzle- I needed the DP portion to start before I could solve it. Thanks for providing such good content and a mix of easier and more difficult problems.
This seems to be one of a number of similar problems that are special situations. We are given a 45 degree ray from the midpoint D of AB and an intersecting ray from A to C. The problem is to find angle ACD. The method of solution only works if the point C is chosen so that angle DCB is 15 degrees or alternatively angle DBC is 30 degrees..
an easier solution i think would be to just draw a line say CE parallel to AD now using alternate interior angle property we can conclude that angle ADC = angle DCE. Now since the line CE is also parallel to DB therefore we can conclude using alternate angle property that angle DBC = angle BCE. Now angle DCE= angle DCB + angle BCE = 45. Therefore, 15 + BCE = 45, therefore BCE = DBC = 30
@@Z7youtube It's in my comment. Using formula for sine of a difference. sin(135°-x) = sin(135°)cos(x)-cos(135°)sin(x) = (√2/2) cos(x) + (√2/2)sin(x) We used the fact sin(135°)=sin(45°)=√2/2 and cos(135°)=-cos(45°)=-√2/2
I admit I got stuck after step one. designating point P was a clever trick, but only worked out by chance, due to the particular properties of the two triangles. After it turned out that the vertex in D of APD is 60 degrees, the rest was easy!
In 1970, my Maths teacher at Whitchurch high school in Cardiff, Mr Smth, was so good he helped me to learn how to do this stuff and I am forever grateful. I still haven't found a use for the integration of 1 + Tan squared (x) though. :)
I think the question may be solved by applying basic triangle rules. I tried by following method- 1. By external angle theorem: angle DBC = 30° 2. In triangle DBC, if length of DB is 'a' (against 15°), then CD is 2a (against 30°) 3. Now, in triangle CAD, AD = a, CD = 2a 4. Angle ACD = x, Angle ADC = 45° = y, Angle CAD = z 5. Angle z = 2 (Angle x) [because CD = 2 (AD)]. 6. As, Angle x + y + z = 180° x + 45 + 2(x) = 180° 3x = 180° - 45° = 135° x = 135°/3 x = 45°
There is no rule in a triangle as you wrote "if length of DB is 'a' (against 15°), then CD is 2a (against 30°)". By this info you can just write: CD>AB.
Question? I am very rusty in terms of geometry. The last course taken was during the 1984-1985 academic school year. From the diagram above might we say that 180 = 30 + ( x + 15 ) + Angle CAD and 180 = 45 + x + Angle CAD and we know by the rule of supplementary angles 180 = 45 + Angle CDB. So, by this rule, we know that Angle CDB has a value of 135. And, by the exterior angle rule given, 135 = x + Angle CAD. Based upon this information, drawn from the given, is it possible that angle x equals 25 and Angle CAD equals 110. Those numbers might be arising in my vision from the CBD smoke seeping in through the walls in this apartment unit; because, I do not smoke and do not physically tolerate it well. A year or so after taking geometry, linear algebra taught that based upon the information given within a problems statement more than one "answer" might exist in the solution-space. And, that might be what is occurring here. If, I am not mistaken. So, we both might have settled on accurate solutions. I still a little puzzled by that external angle rule. Unless, Angle CAD can also have a possible measure of 105. Maybe, that is why it is labelled CAD?😄
During my jee preparation (engineering entrance exam of India) i studied a theorem called m-n cot theorem, and really it gave the answer in just 5-10 seconds :) 😁 BTW great explanation sir😊
All angles except ones on left side are known. The ones on left side are x and 135-x. Use law of sines to relate ratio of shared side length to length of congruent side. Solve for x. x = invtan(sin135(sin30/sin15+cos135)^-1) = 30.
THE SHORT: If you could draw DP so that DP=DB, on paper, you could also just as easily draw AP so that AP=AD=CP=PD, on paper... all anchored by the given angles of ADC & DCB. THE LONG: Imagine vertices ABC are 3 stars in a system, with D being a star exactly between star A and star B (let's say with Hubble, we're easily able to measure that the star D is exactly half way, in light years, between stars A & B, all three nicely aligned). From ANY of these given stars, to any other star (in the ABCD set of star system), as well as, angles 15 & 45....nothing was given, could have been given, via assumption or "eye balling".... Everything came about through precise measurements via parallax + trigonometric, standard Candle method, etc. Take one step back and ask yourself: --- what is the astronomical (if we took ABCD to be a set of 4 stars in the Milky Way) or geometric logic that said you could, for example, draw a precisely KNOWN line segment, like DP, from one vertex of BDC, to the opposite side, BC, in such a way that DP is exactly equal to one side, DB.... but NOT be able to do a similar thing, say, AP? After all, AP=AD=DB, DB=DP, and AD=DB.... all anchored by the given angles of 15 (DCB) & 45 (ADC). For example, "P" can NOT be in any other place, along BC, such that AD is not equal to AP (due to the dictates of the angles 15 and 45, as given). In other words, there's neither "wiggle room" for angles ADC & DCB, on the one hand, nor for point "P" along BC such that AP=AD=DB=DP is NOT true, on the other.
Notice that sin(x)/AD = sin(45)/AC and sin(x+15)/(2*AD) = sin(30)/AC. Combine to get 2*sin(x)/sin(x+15) = sin(45)/sin(30). Upon expansion of sin(x+15) and some algebra, we get x = 30 degrees.
Great puzzle! I didn't find this excellent geometry solution. Instead I dropped a perpendicular from C to point E, then used trig to get CE, EA and EB (after setting AD and BD=1). Then I used Pythagoras to get CB=sqrt 2. Then the Law of Sines to get sin x=1/2. Thanks PreMath!
@@hussainfawzer Hello Thanks for the question, I left out a lot of steps and also erred in saying CB=sqrt2. Sorry for that (CA=sqrt2 is correct). To get EA then CE let EA=y and AB=1 (or any number but 1 makes the calculating easier) Since triangle CED is a right triangle with a 45 degree angle, CE=ED=1+y. EB also =2+y Since angle CEB is defined as a right angle and CBE (as shown by PreMath) is 30 degrees, CEB is a 30-60-90 triangle and CB will be twice CE or 2+2y So by Pythagoras CE^2+EB^2=CB^2 or (1+y)^2+(2+y)^2=(2+2y)^2 Simplify and get 2y^2+2y-1=0 and so y=(sqrt3-1)/2 and EC=(sqrt3+1)/2 I hope that answers your question, Hussain. Now one can get CA=sqrt2 from Pythagoras. And looking at triangle CAD, the law of sines gives sinx/1=sin45/sqrt2 and sinx=1/2.
@@waheisel That was beautiful math. It reminded me of a lot of stuff. Ty I used to love math and physics with a dream to be get into cosmology. Due to personal reasons I'm now soon becoming a doctor. It's good but I miss math and physics so badly 😅
@@yuda4626 Thanks for the kind comment Ty. When I was in school I also liked math, physics, and astronomy. I wasn't nearly good enough to make any of those a career. I also became a doctor. Even though you won't use so much math and physics in your career you can still enjoy your daily PreMath puzzle! Best Wishes
I took Geo Trig last school year and felt very smart knowing how to solve everything. However when I saw this, it was not the case. Thanks for explaining this very thoroughly. It was very fun and interesting to learn how to solve this Geo Challenge. Please make more!
After watching few similar videos able to solve this one in mind. It's mostly in finding/drawing the correct isosceles, equilateral or congruent triangles with the equal length sides.
I got it. My sister told me the triangle CDP is an isosceles because angle PCD and PDC have the same 15 degree. Therefore, PC = PD. The triangle APD is equilateral, hence PC = PA. So angle PCA and PAC are 45 degree each. Consequently, angle ACD is 30 degree. That’s awesome!!! Thank you very much. I had learned geometry 50 years ago. I have to confess this brings back a lot of memories.
I did it without construction. Just let the two equal sides AD = DB = a. And also, let AC = y. That means AB = 2a. Angle B = 30' (Ext. angle of a triangle) In triangle ACD, using the sine rule, we have: sinx/a = sin45/y Thus, y = a.sin45/sinx . . . . . (1) Similarly, in triangle ACB, using the sine rule, we have: sin(x+15)/2a = sin30/y Thus y = 2a. sin30/sin(x+15) . . . . ....(2) From (1) and (2): a. sin45/sinx = 2a. sin30/sin(x+15) sin45/sinx = 1/sin(x+15) sin45. sin(x+15) = sinx . . . . . .(3) Recall the compound angle: sin(A+B) = sinAcosB + cosAsinB (3) becomes: sin(x+15) = sinx/sin45 sinx. cos15 + cosx. sin15 = sinx/sin45 . . . [Divide throughout by sinx] sinx/sinx. cos15 + cosx/sinx . sin15 = 1/sin15 cos15 + sin15/tanx = 1/sin15 sin15/tanx = [1/sin15] - cos15 tanx = sin15/[(1/sin15) - cos15] x = arctan{sin15/[(1/sin15) - cos15]} x = 30
Beautiful solution to the problem. Very much in the domain where math is art itself. I have opted for a cruder way (PS: I am an engineer). I dropped a perpendicular instead and used expressions for tan. Got two equations and soleved simultaneously to get the answer.
I was just wondering how will you use that information, that those two parts of bottom line is equal, and I'm speechless how beautifully you created a triangle out of it and used that fact... I knew that equal sides of triangle subtend equal opposite angles in a triangle, but i wasn't smart enough to use that fact... Thank you sir, love from India..!
Is it only me who thought that if AD=DB, CD is the median line of the triangle in C, thus meaning in cuts the angle in half? And if DCB is 15, ACB would be 30?🤔🤔
Actually, that is not always the case, as shown in this problem here. The median of a triangle in a given vertex is not always equal to the bisector of the relative angle. They're the same in some cases, for example if the triangle is equilateral, it is true for alla the vertices.
You're mistaken between median line and angle bisector. This is proven false in this example so if it cit ACB in middle = ACD = DCB =15° which is not the case
Overly and unnecessarily long for a simple double angle problem by inspection. There are two overlapping triangles, with a common side. The length of which is irrelevant when you apply the sin law and end up with Sin(x+15)/Sin(x) = sqrt(2/3). Basic trig on the 30/60/90 triangle has Sin(x+15)/Sin(x) = Sin45/Sin30. Thus x=30deg is the only viable solution.
Hi, you could have separated triangle cdb and joined AD and DB by rotating clock wise triangle CDB. Making points A and B to be the same point. And you can get the answer in 2 steps. As rotating would create an isosceles triangle. Try it out
Rotating does not create an isosceles triangle. In fact, in the general case, the rotation is more likely to create a 4 sided polygon. In this particular case it creates a triangle, but not an isosceles one. To think this through, think about point C being split by the rotation, where C1 is the original point C, and C2 is created by the rotation. In this case, because angle ADC is 45 degrees and angle CDB is 135 degrees, after the rotation the line segments C1D and DC2 are collinear forming a straight line, with D in the center, since C1D and DC2 are the same length (but since they form a straight line they are not two sides of an isosceles triangle). So one side of the resulting triangle is C1C2, with the other two sides being AC1 and BC2 (A and B are the same point after the rotation). One of the angles in this new triangle is the 15 degrees from angle DCB prior to rotation, another angle is X degrees, formed from angle ACD prior to rotation (now AC1C2 after rotation). We know from the solution presented that X is not 15 degrees but is a value (trying not to spoil) such that none of the angles in the triangle formed by the rotation are the same.
You could have also used this formula, it would have made the question simple (m+n) Cot θ = m Cot α - n Cot β Now as the side is divided into equal ratios, thus m=n=1 Also θ=135°, α=x, β=15° After putting the respective values, (1+1) Cot 135° = (1) Cot x - (1) Cot 15° 2 Cot 135° = Cot x - Cot 15° Cot x = 2 Cot 135° + Cot 15° Cot x = 2(-1) + (2 + √3) Cot x = -2 + 2 +√3 Cot x = √3 x = 30°
i guess in these kind of problems only the one who created them can solve them, or you can go through a long journey of trying all possible methods and theoremes to finally conclude it
on triangle CDB: angle B = 180-(15-(180-45)) = 30 let AD=DB=a, AC=b on triangle ABC, according to the law of sines: 2a/sin(x+15) = b/sin(30) = 2b a/b = sin(x+15) on triangle ACD, according to the law of sines: a/sin(x) = b/sin(45) = b*sqrt(2) a/b = sin(x)*sqrt(2) now we have a single equation to work with: sqrt(2)*sin(x) = sin(x+15) according to the sum of sines identity: sqrt(2)*sin(x) = sin(x)*cos(15)+sin(15)*cos(x) (sqrt(2)-cos(15))*sin(x) = sin(15)*cos(x) tan(x) = sin(15)/(sqrt(2)-cos(15)) = 1/sqrt(3) x=30
One way I solved this problem was using the law of sines, such that for side lengths A, B, and C and respective opposing angles a, b, and c we have A/sin(a) = B/sin(b) = C/sin(c). After finding angle CBA through Exterior Angle Theorem, we can denote the length of AC as y and BD/AD as s (the variables are arbitrary and will eventually cancel out) and we get two equations. y/sin(45) = s/sin(x) and y/sin(30) = 2s/sin(x+15). Solving for y for the first equation we get y = s*sin(45)/sin(x), which we can plug into the second equation to get s*sin(45)/sin(x)sin(30) = 2s/sin(x+15). We can factor out the s and put all x values on one side to get sin(x+15)/sin(x) = 2sin(30)/sin(45). Now, having sin(x+15) on the top is important, as we use the Compound Angle Formula for Sines (which is sin(a+b) = sin(a)cos(b) + sin(b)cos(a)) to make the left side (sin(x)cos(15) + sin(15)cos(x))/sin(x) and by the rule of common denominators that becomes (sin(x)cos(15))/sin(x) + (sin(15)cos(x))/sin(x), and by doing some factor reduction we have the equation become sin(15)*cot(x) + cos(15) = 2sin(30)/sin(45), which becomes cot(x) = (2sin(30)/sin(45) - cos(15))/sin(15) when you isolate x. Do a bit of calculations and taking the inverse cotangent and you end up getting 30 degrees.
I solve this with the theorem of Euler. After I find B angle 30 degrees we applied the Theorem and with intersting goniometric equation we can find X value
Me too. Then, as tan 15°=2 - √3 (use the compound angle formula to evaluate tan 15°=tan(45°-30°) to get this), we get x=45°-15°=30°. I didn't spot the geometric solution until I saw the construction of AP in the video.
@@PreMath And here the details for all others: Draw the height from point C to the A-B line. Call the foot point as E. In my way, I also calculated the angle at point B first, i.e. 30°. The triangle BCE shows a 30°/60°/90° triangle. So when BC has a length of 2 units, then CE has lenght 1 unit and BE is sqrt(3) units long. CD then is sqrt(2) units long because of the 45° angle and DE is 1 unit long (40°/45°/90° triangle CDE). Lengt BD = BE - DE = sqrt(3) - 1. Because AD = BD =sqrt(3)-1 units then AE = DE - AD = 1 - (sqrt(3) - 1) = 1 - sqrt(3) + 1 = 2 - sqrt(3). Focus on the right triangle ACE where the angle ACE can be calculated as tan(ACE) = AE/CD = (2 - sqrt(3))/1 = 2 - sqrt(3). And important to know: tan(15°)=2-sqrt(3) is also a special value (can be easily proved on a 75°/75°/30° isosceles triangle). The actual result x is just the difference of the angle BCE (45°) and ACE (15°) in this case.
Another method is to use sine theorem and a trig identity to prove sin x /cos x = (sqrt 3)/3 = tan x. So x =30. Less head scratching to figure out how to draw auxillary lines.
Generality is not lost even if AD=BD=1 ∴CD=sin30°/sin15°=2cos15° From the sine theorem for △ACD 2cos15°/sin(135°-x)=1/sinx ∴ 2cos15°sinx=sin(135°-x) ∴ sin(x+15°)+sin(x-15°)=sin(45°+x) ∴ sin(x+15°)=sin(45°+x)-sin(x-15°)=2cos(x+15°)sin30°=sin(75°-x) ∴ 2x=60° ∴ x=30°
Answer x=30 degree I did something different using the 30-60,-90 right triangle which has the sides 1, 2, and sqrt 3 (sqrt 3=1.732) and the 45-45-90 right triangle. Construct a line perpendicular from C to form 30, 60, 90 triangles CPA distance CP hence =1 PA= sqrt 3 and CA= 2 (the hypotenuse) let's label angle CPA as 'y' and y+x=45 degree. you now as 45-90-45 right triangle with sides 1 1 and sqrt 2. PD =1 and CD sqrt 2 Using ASA (15 degrees, sqrt 2, and 135 degrees) produces 0.732 for DB, but DB=AD thus PA =0.268 (1-0.732). Right triangle CPA now has sides 1, 0.268, and 90 degrees, and using SAS gives 15 degrees for y. but y+x =45 degree hence x= 45-15 =30 degree answer
From law of sines i got that x=30° From law of sines I got a/c=sin(15°)/sin(30°) (in triangle BDC) and a/c=sin(x)/sin(135°-x) (in triangle DAC) Comparing this two i got sin(x)/sin(135°-x)=sin(15°)/sin(30°) which lead us to sin(x)/(cos(x)+sin(x))=1/2*(sqrt(3)-1) tan(x)/(1+tan(x))=1/2*(sqrt(3)-1) t/(1+t)=1/2*(sqrt(3)-1) 2t/(1+t)=(sqrt(3)-1) 2t=(sqrt(3)-1)(1+t) 2t=(sqrt(3)-1)+(sqrt(3)-1)t 2t-(sqrt(3)-1)t=(sqrt(3)-1) (3-sqrt(3))t=(sqrt(3)-1) t = (sqrt(3)-1)/(3-sqrt(3)) t=1/6(sqrt(3)-1)(3+sqrt(3)) t=1/6(3sqrt(3)+3-3-sqrt(3)) t=1/6*2sqrt(3) tan(x)=sqrt(3)/3 and in the interval [0,180°] only x=30° satisfies equation tan(x)=sqrt(3)/3
I used sin rule. Sin15/DB = sin30/CD. I get DB = 0.518CD. and then sinx/DB = sin(135-x)/CD. I substituted DB = 0.518CD into the second sin rule equation. CD got cancelled out and after using some trigo identities, I managed to solve for x as the answer given. Good qns!
@@starpupil1843 u r welcome! My weakness involves me not knowing what line to draw or what to substitute (in calculus) to get to my solution faster. I deal with what I see and what I can find base on what I see. Hahs
Extend line CA and obtain exterior angle (180-x-15) =x+ ang CAD thats one equation. Another eqn is using triang.CAD : x + Ang CAD =135. We have two eqns and two unknowns which we can solve to get x=30
A rectangle is completed by dropping a perimeter C to "F" and with segment "F" to A completes the triangle with right angle at "F". A line segment C to "E" drops a line to B thus forming the triangle C"E"B so completing the rectangle as such; the assumption being that any rectangle that is "ideal" is also a rectangle that is derived from the given relations of the known angles to the nominal rectangle alone. With the angles as given it is a simple process to account for respective right angles and so derive the triangle "C F A" with angles 15 90 and 75 degrees. These values do not require trigonometric operations since they are self evident from the given angles and the fact that a perimeter line segment with points "a" and "b" , a bearing and azimuth, confirms the angle value of any single point as the difference value of the opposite point. In other words, based on which direction (point) one initially chooses. Whether or not this constitutes a valid proof, it is far easier and intuitive. I simply noted the assumed initial line segment relations to angles given and derived x= 30 degrees accordingly.
Nice proof. I have one question remaining though: how do you justify the existence of the point P without referencing to the picture of the triangle ABC?
Using the law of sines 1st .......for both ACD and BCD triangles.........you will get 2 equations one relating H and M from BCD triangle(H represents both the AD and BD lines,as they are the same and M represents the CD line).........the other relating H,M and x(the angle)...........the solution of 2 equations will lead to the value of x which turns out to be 30°
1/ Just build a right special 60-90-30 triangle APB as in the picture. we have the angle APB= 90 degrees, D is the midpoint of AB. we have PA=PD=AD=DB so the triangle APD is a equilateral, and PDB is an isosceles of which the 2 bases angle value= 30 degrees. 2/ Now extend PB a segment PC such that PC=PA, we have: the triangle APC is a right isosceles so the angle ACP= 45 degrees; CPD is an isosceles so the angle DCP= 15 degrees( half of DPB); and the angle ADC=60-15= 45 degrees. x= 45-15 = 30 degrees. The problem is solved inversely.
I found the value of x by a complete differente way 0-0 Using the sinus relation: a/sin a = b/sin b = c/sin c And the Addition Sinus formula Sin(a+b) = sin a*cos b + sin b* cos a
Japanese math education teaches students to find similarity triangles in the figure. then find correlate angles. in this case, angle 45degree is equal to angle C. THEN, x =45-15.
Draw PD = BD = AD So angle PDA =(angle PDB +angle PBD)=60° So ∆ PDA is equilateral. And angle CPA = angle PAD +angle PBD =90° Again angle PDC =angle DPB - angle PCD =15° Hereby CP= PD = AP So, CPA is right angular isosceles. Hereby x = angle ACP - 15° = 30°
Let |ad|=|db|=1 unit.Using the sine rule find |cd| in the triangle cdb. In the triangle ACD you know |cd| and |ad| and the angle 45 degrees. Using the cosine rule find|ac|. Then use sine rule to find x.
Я решил по-другому, не проще - скорее сложнее. Если вокруг BDC описать окружность, то в ней BC - сторона вписанного квадрата. => аналогичный сегмент получается поворотом на 90 градусов. Если продлить "повернутую" сторону B'D' до пересечения с BD в точке M, то CMB - прямой угол. Я не буду приводить детали, там любой легко выразит CM=MD и DB через, например, радиус окружности. А заодно легко найти AC из треугольника ACM. После этого легко показать, что степень точки A AB*AD = AC², то есть AC - касательная к окружности (BDC). => ∠ACD = ∠DBC;
I remember a short cut, but I'm old and don't know where I learned it. Basically if the line segments are the same then the angle x = the angle CBD. so all you have to do is solver for CBD. Angle ADC may have to be 45 though(it might have to do with trig). I don't remember.
Let L=AD=DB and S=AC By exterior angle formula angle DBC=30d By Sine Rule on triangle ACB and then triangle DBC: SIN(X+15)=(2L/S)SIN(30)=L/S=SIN(X)/SIN(45)=(^/2)SIN(X) Therefore X=30 as Sin(30+15)=Sin(45)=1/(^/2)=(^/2)/(2)=(^/2)Sin(30)
