Can you find area of the Triangle? | (Simple Rule) | 

Thanks Sir for your solution, quite different than usual, but efficient

BD=x DC=√3x AB+BD=DC 2+x=√3x 　　　2=(√3-1)x x=2/(√3-1)=√3+1 DC=3+√3 area of the Triangle : 2*(3+√3)*1/2=3+√3

Let the corner of the right angle be D, and let BD be "a". h = 2+a = a(sqrt3). a = 1 + sqrt3. h = 3 + sqrt3. A = 0.5 x 2 x h = 3 + sqrt3.

Draw CD perpendicular to AB produced. BD = x, BC = 2x, CD = 2+x. Use Pythagoras theorem and find x . X= 1+ root 3. Area of ABC = 1/2(3+roor3)×2=3+root3

BD Right CD CAD=45°=ACD So BCF=30° In ∆ BCD (30°;60°;90°) Let BD=x ; BC=2x; x√3 AD=AB+BD=2+x DAC=ACD=45° So 2+x=x√3 x=√3+1 So CD=√3(√3+1)=3+√3 So area of the triangle=1/2(2)(3+√3)=3+√3 square units=4.73 square units.❤❤❤

Extend AB to D and drop a perpendicular from C to D. As ∠CAD = 45° and ∠ADC = 90°, ∠ECA = 90°-45° = 45°. As ∠CAE = ∠ECA = 45°, ∆AEC is an isosceles right triangle and EC = AE. As ∠ECA = 45° and ∠BCA = 15°, ∠ECB = 45°-15° = 30°. As ∠BEC = 90°, ∠CBE = 90°-30° = 60°, and ∆BEC is a 30-60-90 special right triangle. This means that CB = 2BE and EC = √3BE. Let BE = x. AE = EC AB + BE = EC 2 + x = √3x √3x - x = 2 x(√3-1) = 2 x = 2/(√3-1) x = 2(√3+1)/(√3-1)(√3+1) x = 2(√3+1)/(3-1) x = 2(√3+1)/2 = √3 + 1 EC = √3x = √3(√3+1) = 3 + √3 Triangle ∆ABC: Aᴛ = bh/2 = 2(3+√3)/2 = 3 + √3 ≈ 4.732 sq units

as other viewers pointed, it can be solved without trigonometry, wich I find it easier. thank you for the video!

Let's find the area with a simple method: . .. ... .... ..... Let's add point D such that B is located on AD and ACD is a right triangle. From the known angles we obtain: ∠ABC = 180° − ∠BAC − ∠ACB = 180° − 45° − 15° = 120° ∠DBC = 180° − ∠ABC = 180° − 120° = 60° Since ∠BDC=90°, the triangle BCD is a 30°-60°-90° triangle. Therefore we know: BD:CD:BC = 1:√3:2 ⇒ BD/CD = 1/√3 Additionally we know that ∠ACD=∠ACB+∠BCD=15°+30°=45°=∠BAC=∠DAC. Therefore the triangle ACD is an isosceles right triangle and we can conclude: CD = AD BD/CD = 1/√3 BD/AD = 1/√3 (AD − AB)/AD = 1/√3 (AD − 2)/AD = 1/√3 √3*AD − 2√3 = AD (√3 − 1)*AD = 2√3 ⇒ CD = AD = 2√3/(√3 − 1) = 2√3(√3 + 1)/[(√3 − 1)(√3 + 1)] = 2√3(√3 + 1)/(3 − 1) = √3(√3 + 1) = 3 + √3 CD is the height of the triangle ABC according to its base AB. So we can finally calculate the area of the triangle ABC: A(ABC) = (1/2)*AB*h(AB) = (1/2)*AB*CD = (1/2)*2*(3 + √3) = 3 + √3 Best regards from Germany

Solution: a√3 = 2 + a a√3 - a = 2 a(√3 - 1) = 2 a = 2/(√3 -1) a = 2√3 + 2/2 *a = √3 + 1* AD = DC 2 + (√3 + 1) = DC DC = √3 + 3 A = ½ 2 (√3 + 3) A = √3 + 3 Square Units A ~= 4,732 Square Units

Let base of dropped perpendicular h be point D. Then triangle ADC is iscoseles 45, 45, 90 degree. So AD = DC. Thus BD = h - 2. Triangle BDC. Tan 60 = h / h - 2. 1.732 = h / h - 2. 1.732h - 3.464 = h. 0.732h = 3.464. h = 4.732. Area = 1/2 * 2 * 4.732. 4.732.

The fun begins when Trigonometric Formulas are divided into 2 major systems: ●Trigonometric Identities ●Trigonometric Ratios ☠ Caution Triangulation Formulas are the leading cause of Strangulation. 🙂

4.732051 A different approach Made labeling error by using 2 (instead of sqrt 2) but will use the same approach Draw a 45-degree line from B to form a 45-45-90 right triangle ABP Since, the hypotenuse = 2, and sine in a 45-45-90 triangle the sides are x, x and x sqrt 2 then 2 = x sqrt 2 2/sqrt2 =x sqrt 2 = x Hence, BP= sqrt 2 and AP= sqrt 2. Hence, the area of triangle ABP= sqrt 2* sqrt 2*1/2 = 1 AREA OF ABP Let's find the area of triangle BCP. For triangle BCP (which was formed when triangle ABP was formed), which is a 15, 75, and 90-degree right triangle (since angle B = 120 and 120-45=7), In a 15-75-90 right triangle, the ratio of the angles are sqrt 3-1, sqrt 3 + 1, and 2 sqrt 2 The 15-degree angle corresponds to sqrt 3-1 or 0.732051 and the 75-degree angle corresponds to sqrt 3+1 2.732051 Since in triangle BCP, BP is opposite 15 degrees, and BP length is sqrt 2 (the value for x) then CP (which is opposite the 75 degrees) can be found by dividing sqrt 2 by 0.732051, then multiply the result by 2.73205 Hence, sqrt 2/0.73205 * 2.732051 or sqrt 2/ sqrt 3-1 * sqrt 3+ 1 1.93815114 * 2732051 BP= 5.27791 AP= sqrt 2 Area of BCP = 5.27791 * sqrt 2 * 1/2 = 3.73205 Recall area of ABP = 1 Hence, the area of both or the triangle above = 1 + 3.72305 = 4.72305. Answer ------------------------------------------------------------------------------------------------ The area of a 75-15-90 right triangle can also be found by knowing the length of the hypotenuse. recall that 90 degrees correspond to 2 sqrt 2 So divide sqrt 2 / 0.73205 = 1.93815114 then multiply the result by 2sqrt 2 = 5.4641 In 15, 75, and 90 degrees, the area is = the hypotenuse square divided by 8 Hence, 5.4641^2/8 = 29.85639= 3.73205, then add the 1 ( the area of ABP) = 4.73205 again the same answer

The Answer is (3 + sqrt(3)) Square Units. Why? Because the Height of the Triangle [AOC] = (h) = (2 + a) is equal to the Base of an Isosceles Triangle [AOC] with Angles (45º ; 45º ; 90º). tan(45º) = 1. As : tan(60º) = sqrt(3) then : tan(60º) = (a + 2) / a So : (a + 2) / a = sqrt(3) Single Solution : a = (sqrt(3) + 3) Base = 2 height = (sqrt(3) + 3) Area = (sqrt(3) + 3) sq un And that's all folks.

Drop a perpendicular from B to AC and label the intersection as point D. ΔABD is an isosceles right triangle. AD and BD have lengths √2. ΔBCD has a given 15° angle,

My way of solution ▶ we have a triangle Δ ABC if we draw the height of this triangle, to point D, we would have a right triangle: ΔADC there is also a smal right triangle ΔDBC, in this triangle ΔADC. [AB]= 2 [BD]= x let's consider the triangle ΔBDC [BD]= x ∠CBD= 60° ⇒ tan(60°)= [DC]/[BD] √3= [DC]/x [DC]= √3x Let's consider the right triangle ΔADC: sin(45°)= [DC]/[CA] √2/2= √3x / [CA] [CA]= √6x according to the Pythagorean theorem: [AD]²+[DC]²= [CA]² [AD]= 2+x [DC]= √3x [CA]= √6x ⇒ (2+x)²+(√3x)²= (√6x)² 4+4x+x²+3x²= 6x² 2x²-4x-4=0 x²-2x-2=0 Δ= 4+4*1*2 Δ= 12 √Δ= 2√3 x₁= (2+ 2√3)/2 x₁= 1+ √3 x₂= (2-2√3)/2 x₂= 1-√3 x₂ < 0 ❌ ⇒ x= 1+ √3 ⇒ [CA]= √6x [CA]= √6*(1+ √3) [CA]= √6+ 3√2 A(ΔABC)= 1/2* [CA]*[AB]*sin(45°) A(ΔABC)= 1/2* (√6+ 3√2)*2*√2/2 A(ΔABC)= 3+√3 A(ΔABC)= 4,732 square units

Other method: H the orthogonal projection of C on (AB) and x = BH AHC is right isosceles and AH = HC = 2 + x In triangle HBC: tan(angleHBC) = tan(60°) = sqrt(3) = HC/BC, so sqrt(3) = (2 + x)/x, giving x = 1 + sqrt(3) when simplified. So HC = 2 + x = 3 + sqrt(3) and the area of ABC is (1/2).AB.BC = (1/2).(2).(3 +sqrt(3)) = 3 +sqrt(3).

When sine and cosine laws are not to be used, how can their derivative tan be used?

Super💡

Drop a line from B to AC in D such as the line is perpendicular to AB ABD is a isocèles right triangle so BD= 2 AD= 2*sqrt 2 by the Pythagorean theorem Angle BDC is 180-45=135 So Angle CBD is 15 degrees So we get that BDC is isoceles DC=BD=2 AC=2(1+sqrt2) AB=2 Area of the triangle 1/2 ab sin c 1/2*2*2(1+sqrt2)*sin 45 =sqrt 2+2

Nice! φ = 30° → sin⁡(3φ) = 1; sin⁡(φ) = 1/2 → cos⁡(φ) = √3/2 → sin⁡(φ/2) = √((1/2)(1 - cos⁡(φ))) = (√2/4)(√3 - 1) sin⁡(4φ) = sin⁡(6φ - 4φ) = sin⁡(2φ) = cos⁡(φ) = √3/2 ∆ ABC → AB = 2; BC = k; AC = AT + CT = AT + a; CAB = 3φ/2; BCA = φ/2 → ABC = 4φ sin⁡(BTA) = 1 → AT = BT = √2 → sin⁡(φ/2) = (√2/4)(√3 - 1) = √2/k → k = 2(√3 + 1) → area ∆ ABC = (1/2)sin⁡(4φ)2k = (√3)(√3 + 1) or: φ = 30°; AO = AB + BO = 2 + BO; BC = k; CO = AO; OCB = φ → BO = k/2 → CO = k√3/2 = 2 + k/2 → k = 2(√3 + 1) → 2 + k/2 = 3 + √3 = AO = CO → (1/2)2(3 + √3) = (√3)(√3 +1) = area ∆ ABC

2+x=sqrt(3)x, x=2/(sqrt(3)-1), then the area is 1/2×2×sqrt(3)×(2/(sqrt(3)-1))=sqrt(3)×(sqrt(3)+1)=3+sqrt(3).😊

Let's use an orthonormal, center A, first axis (AB) Equation of (AC): y = x Equation of (BC): y = tan(60°).(x - 2) = sqrt(3).x - 2.sqrt(3) At the intersection (point C) xe have: (sqrt(3) -1).x = 2.sqrt(3), so x = 3 + 2.sqrt(3) when simplified. The ordinate of C is the same as its abscissa, it is the height of the triangle ABC: 3 + sqrt(3), when its basis is AB = 2 So the area of the triangle is (1/2).(2).(3 + sqrt(3)). It is then 3 + sqrt(3).

2+x=√3x x=2/(√3-1) x=√3+1 AC=√2(3+√3) Area= ½(2)√2(3+√3)√2/2 Area=3+√3

u can reflect trianle across AC to form right angle triangle, it will give 30 60 90 trianglee, rest is easy

Inscribimos el triángulo ABC en el cuadrado ADCE de lado AD=a y cuya diagonal es AC→ El triángulo BDC tiene ángulos 60º/90º/30º→ BD*√3=DC→ (a-2)√3=a→ a=3+√3→ Área ABC=AB*DC/2=2*a/2=a =3+√3. Gracias y un saludo cordial.

Posti a,b i due black Lines,risulta asin45=bcos30..a=√3b/√2..acos45-bsin30=2..√2a/2=b/2+2..b=4/(√3-1)=2(√3+1)..a=√6(√3+1)..A=(1/2)absin15

i suppose we need the perpendicular height. Make a point D to the right of AB, such that ADC is a right triangle. AD is x + 2

I have had a little look, and I found the formula h = x/(cot alpha - cot beta). Looks simpler than yours frankly. And it works!

S=3+√3≈4,732

Where did you have the formula to find h?

Thank you

1:51 Do you have a proof for this formula for h? (I note that it is the general answer to the surveyor's problem of determining a height when you can't explicitly measure the distance to the base of the object being measured.)

Are alpha and beta can be any angle or some specific angle ???? Pls reply

Where that formula comes???

This is a great demonstration of how to solve using law of tangents. It seems that your intent has escaped some grumpy commenters. 😂

Using tan but banning sine and cosine makes zero sense. There should be no trigonometry allowed if you are gonna play that game. And this problem can be solved without the trigonometric functions.