Can you solve without using Trigonometry? 

The ratio of sides of a 22.5°-67.5°-90° right triangle is derived from constructions in a regular octagon, applying the properties of an isosceles right triangle, the Pythagorean theorem and additional basic geometry, but trigonometry need not be used. The ratio (short side):(long side):(hypotenuse) is 1:(1 + √2):√(4 + 2√2). Extend AB down and drop a perpendicular from C to it, labelling the intersection as point D. ΔBCD is a 22.5°-67.5°-90° right triangle and ΔACD is an isosceles right triangle. Let BD = x. Then CD = x(1 + √2). By properties of an isosceles triangle, AD = CD = x(1 + √2). AB = AD - BD = x(1 + √2) - x = x√2. Let AB be the base of ΔABC, then CD is the height and the area is (1/2)(x√2)(x)(1 + √2) = x²(2 + √2)/2. We note that the hypotenuse is x√(4 + 2√2) and also, from the given, equal to 4. So, x√(4 + 2√2) = 4 and x = 4/(√(4 + 2√2)). Therefore, x² = (4/(√(4 + 2√2)))² = 16/(4 + 2√2) = 8/(2 + √2) and area = (8/(2 + √2))(2 + √2)/2 = 4, as Math Booster also found.

4 From B, Draw a 22.5-degree angle to touch AC. Let's label the point P BPC = 135 degrees Hence, triangle ABP is a 45, 45, 90 right triangle. Let's label BP = n Then AB = n and PC =n Draw a perpendicular line from P to BC to create two congruent triangles ( 67.5 degrees 90 degrees and 22.5 degrees) BPR and CPR CR= 2 = BR Notice that in both BPR and CPR, 2 is opposite 67.5 degrees (I will come back to this later) with n as their hypotenuse Drop a line from A to X to form a right triangle ABX This is similar to BPR and CPR (67.5 degrees, 90 degrees, and 22.5 degrees In triangle, ABX n is also it's hypotenuse and length AX = 2 The area of the triangle is length BC * AC *1/2 4 * 2 *1/2 = 4 Answer

14:00 sin (90+a)=-cos a

Your second method used trigonometry !

The octagon analogy is apt and also a teaching moment. It should lead to discussions about regular polygons including the pentagon, hexagon and decagon..Thats what i did for many years.

good question, thank you😊

First method (no trig): Draw a line segment from B to D on AC so that ∠ABD = 90°. ∠BDA = 90°-45° = 45°, so ∆ABD is an isosceles right triangle and AB = BD. As ∠BDA = 45°, and is an external angle to ∆CDB at D, ∠DBA = 45°-22.5° = 22.5°. Therefore, ∆CDB is an isosceles triangle and DC = BD. Drop a perpendicular from D to E on BC. As ∆CDB is isosceles, and D is the vertex opposite the base, DE bisects ∆CDB and creates two new congruent right triangles, ∆BED and ∆DEC. In right triangle ∆BED, as ∠DBE = 22.5°, ∠ EDB = 90°-22.5° = 67.5°. Extend EB to F and drop a perpendicular from A to F. As ∠FBA is an external angle to ∆ABC at B, ∠FBA = 45°+22.5° = 67.5°. As ∠AFB = 90° and AB = BD, ∆AFB is congruent with ∆BED, and AF = BE = 4/2 = 2. Triangle ∆ABC: A = bh/2 = 4(2)/2 = 4 sq units Second method (trig ok): As ∠A = 45° and ∠C = 22.5°, ∠B = 180°-(45°+22.5°) = 112.5°. sin(22.5°) = sin(45°/2) sin(22.5°) = ±√((1-cos(45°)/2) sin(22.5°) = √((1-1/√2)/2) 0 sin(22.5°) = √((√2-1)/2√2) sin(22.5°) = √(2-√2)/2 sin(112.5°) = sin(180°-112.5°) = sin(67.5°) sin(112.5°) = sin(135°/2) sin(112.5°) = ±√((1-cos(135°)/2) sin(112.5°) = √((1+cos(45°)/2) 0 sin(112.5°) = √((1+1/√2)/2) sin(112.5°) = √((√2+1)/2√2) sin(112.5°) = √(2+√2)/2 By the law of sines: BC/sin(A) = CA/sin(B) = AB/sin(C) 4/sin(45°) = CA/sin(112.5°) = AB/sin(22.5°) 4/(1/√2) = CA/(√(2+√2)/2) = AB/(√(2-√2)/2) 4√2 = 2CA/√(2+√2) = 2AB/√(2-√2) CA = 4√2(√(2+√2))/2 = 2√(4+2√2) AB = 4√2(√(2-√2))/2 = 2√(4-2√2) Area = CA(AB)sin(45°)/2 A = 2√(4+2√2)(2√(4-2√2))/2√2 A = 4√(16-8)/2√2 A = 4√8/2√2 = 8√2/2√2 = 4 sq units

A little *crazy solution* with Pythagoras theorem. I will use Math booster shape in 2:41 Let BE⊥AC (construction) Obviously AE=ED=BE=x and BD=DC=y Pythagoras theorem in Δ BDE => y²=x²+x² => y²=2x² (1) Right triangle BEC (Pythagoras) (x+y)²+x²=16 => 2x²+2xy+y²=16 => 2x²+2xy+⁡2x²=16 cause (1) => *x²+xy+⁡x²=8* (2) Area of triangle ABC (ABC)=(ABE)+(BED)+(BDC) =½x²+½ x² +½•xy =½(x²+x²+xy) =½•8 cause (2) = 4 square units

If trigonometry is permitted, in ΔABC, the length of side AB can be calculated using Law of Sines. Side AB is shared with ΔAPB. The altitude of triangle ABC is the length of side AP of triangle ΔAPB. Since the length of side AB is known, the Law of Sines can again be used to determine the altitude/height of ΔABC which is AP of triangle ΔAPB which is 2. Area = .5 x b x h = .5 x 4 x 2 = 4

Reflect ∆APC about PC and extend AB to AF. ∠PAC = 22.5° and since ∠ACF is 45°, ∆AFC is an Isosceles Right triangle, with AF = FC. ∆AFA' is Congruent to ∆BFC by SAS, hence AA'= 4 AP = ½AA' = ½*4 = 2 Area(∆ABC) = ½*2*4=4

Interesting methods. I have another one. If you draw the line BD to AC to make angle 90 degrees with AC, we can say that: BD = 4*sin22,5, then we can find DC = sqrt(4^2 - (4*sin22,5)^2) = 4*cos22,5. And if angle BAD is equal to 45 degrees then angle ABD is equal to 45 degrees too, then BD = AD. So area of ABC = 1/2*BD*(AD+DC) = 1/2*4*sin22,5*(4*sin22,5+4*cos22,5) = 8*((1-cos45)/2) + 4*sin45 = 4*(1-sqrt(2)/2) + 2*sqrt(2) = 4

Form a right triangle AFC so that ACF is 45 degree and AFC is a right angle. Let AF=CF=a Then AC=a*sqrt(2) By the angle bisector theorem BF=a/(1+sqrt2)=a*(sqrt2 -1) BF^2+FC^2=4^2 by Pythagorian. This results in a^2 = 8 / (2 - sqrt2). So the area of AFC = 1/2 * 8 / (2 - sqrt2) Since the ratio AB/AF = 2 - sqrt2 by angle bisector theorem, the area of ABC = 1/2 * 8 = 4.( 2-sqrt2 on both numerator and denominator cancel each other out)

2nd solution is almost pure trigoninetry based solution. There is another non trigoninetric way, by utilizing Pithagoras theorem in a different set of assisting graphs. one can draw.

Consider a circle through A, B and C with center M. angle(BAC) = 45deg, so angle(BMC)=90deg. angle(ACB)=22,5deg, so angle(AMB)=45deg. Then angle(AMC)=135deg. BM=CM, so angle(MBC)=(180deg-90deg)/2 = 45deg. Then AM parallel to BC, so [ABC] = [BCM] = 4 * 2 / 2 = 4.

this is a teaching moment. The nitti gritti is obvious to everyone who can follow the reasoning. Since we want the area, the first step is to draw the altitude!!When this is not enough, we need to make use of the 45 degree angle.How?Thats the teaching moment.

Both methods are good .Thank you professor.

That is pretty much the third or fourth time that I understood both methods. And looks like I have just found another problem for practice.

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