You don't need the Law of Cosines. Since triangle BEC is isosceles, BE must equal sqrt(3). That makes triangle BEA isosceles, with obtuse angle 150 degrees, so the two small angles are 15 degrees each.

Drop a perpendicular from A to BC and label the intersection as point E. Let AE have length h. We note that ΔACE is a 15°-75°-90° right triangle. Its ratio of sides is (short-long-hypotenuse) (√3 - 1):(√3 + 1):2√2, so CE = (h)(√3 - 1)/( √3 + 1). Multiply by (√3 - 1)/(√3 - 1): CE = (h)(√3 - 1)(√3 - 1)/( √3 + 1)(√3 - 1) = (h)(3 - 2√3 + 1)/(3 - 1) = h(2 - √3). We note that ΔADE is a 30°-60°-90° right triangle. Its ratio of sides is (short-long-hypotenuse) 1:√3:2, so DE = h/√3 = (h√3)/3. So, CD = CE + DE = h(2 - √3) + (h√3)/3 = h(6 - 3√3)/3 + (h√3)/3 = h(6 - 2√3)/3. BD = 1 and CD = 2, so BD is half as long, or h(3 - √3)/3 = h(1 - (√3)/3). BE = BD + DE = h(1 - (√3)/3) + (h√3)/3 = h. So, BE and AE have the same length, h. Therefore, ΔAEB is an isosceles right triangle and

All three sides of triangle ACD can be determined by the law of sines. Since the length of sides BC and AC are [now] known, and the angle ACB [75°], the length of side AB can be determined using the law of cosines. Since the lengths AB and AC are known, as well the angle ACB [75°] the law of sines can be used to calculate θ, 45°

The length of BE is determined once angle EBD is shown to be 30° because triangle BEC has two 30° angles so their opposite sides BE and EC will be the same length.

Very nice problem which posseses a beautiful elementary solution that makes use of no trig, with the only "trick " being used is the 30°-60°-90° triangle, whose side length ratios can be obtained by bisecting an equilateral triangle.

4:06 We can use exterior angle of traingle 5:22 angle CBE=angle BCE=30°.Therefore,BE=EC=√3

As BC is a straight line, the two angles at ∠D must sum to 180°. As ∠ADC = 60°, ∠BDA = 120°. And as ∠BDA is an exterior angle to ∆ADC, it must be equal in value to the two opposite angles in ∆ADC, so ∠CAD = 120-75 = 45°. As ∠CDA is an exterior angle to ∆BDA, ∠DAB = 60°- θ. By the law of sines: AD/sin(75°) = DC/sin(45°) AD/((√6+√2)/4) = 2/(1/√2) AD = 2√2((√6+√2)/4) AD = 4(√3+1)/4 = √3 + 1 Drop a perpendicular from A to E on DC. As ∠E = 90° ∠ADE = 60°, ∠ECA = 75°, and ∠CAD = 45°, ∠EAD = 30° and ∠AEC = 15°. As ∆DEA is a 30-60-90 special right triangle, DE = AD/2 = (√3+1)/2 and EA = (√3)DE = (3+√3)/2. BE = 1 + (√3+1)/2 BE = (2+√3+1)/2 = (3+√3)/2 As EA = BE, ∆BEA is an isosceles right triangle, and ∠EAB = ∠ABE = (180-90)/2 = 45°. So θ = 45°.

Let triangle ABC and AE the height of the triangle. Suppose DE= x . Extend DC and let EF=x . In triangle ADF , AE is height and median at the same time , so ADF is isosceles triangle. In orthogonal triangle ADE DE=x and AD=2x. Pythagoras theorem in ∆ ADE => AE=x•√3 Notice that EC=DC- DE=2-x and CF=DF-CF=2x-2. You can easily prove that AC is angle bisector of EC/CF=AE/AF => (2-x)/(2x-2)=x•√3 / 2x => ……. x =((√3+1)/2 (1) We have BE= BD+DE =1+x =1+ (√3+1)/2 = (3+ √3)/2 = ((√3+1))/2⋅√3 = x√3 = AE So BE=AE …… ∆ ABE is orthogonal and isosceles triangle => θ=45° Have a nice day.

I did it in a much more complicated fashion, by dropping a line from A creating an equilateral triangle, and then using Law of Sines twice and Law of Cosines once to solve for theta. 😅 Hopefully one day I will be like Math Booster.

Using m-n cot theorem solve it instantly

@gekyume9094

2 ай бұрын

Would you be so kind and extrapolate on it?