Can you find area of the triangle? | (Trigonometry) | #math #maths #geometry

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Can you find area of the triangle? | (Trigonometry) | #math #maths #geometry

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[RAG]

I did not use calculator to find x, I used cos x = 5/6, sin x = rt11/6, and angle formulae to find sin 3x = 8rt11/27, giving area is 3380rt11/9. You normally do not calculator for these. You ought to tell us when we can. It makes it so much easier.

[jan-willemreens]

... Good day to you, Step(1) SIN(X)/39 = SIN(2X)/65, where SIN(2X) = 2SIN(X)COS(X) ... after a few algebraic steps ... COS(X) = 5/6 ... X = ARCCOS(5/6) deg. ... 2X = 2ARCCOS(5/6) deg. ... Angle(A) = 180 - 3ARCCOS(5/6) deg. ... Step(2) Finally ... Area(ABC) = (1/2) * 39 * 65 * SIN( 180 - 3ARCCOS(5/6)) = approx. 1245.577 u^2 ... one last remark, I like to approximate only at the very end, not in between! Thanking you for all your instructive math efforts and Happy Wednesday ... best regards, Jan-W

[solimana-soli]

Being 65=13*5 and 39=13*3 we can divide both by 13 to semplify computation.
To find BC we can apply this formula of Brahe:
BC=5cosx + 3cos2x
BC=5cosx + 6cosx^2 - 3
We can find cosx with sines law:
H=5sinx
H=3sin2x
5sinx=6sinx*cosx
Cosx=5/6
BC=5*5/6+6*25/36 - 3 = 16/3
Moltiplying all sides by 3
BC=16, AB=15, AC=9
Applying Heron’s formula with semiperimeter = (16+15+9)/2=20
Area = sqrt 4400=66, 33
Area=66, 33*(1/3^2)*(13^2)=1245, 576…
Or if prefered Area=3380/9sqrt11

[marcgriselhubert]

Let H be the orthogonal projection of A on (BC). In right triangle AHC we have AH = AC.sin(2.x) and in right triangle AHB we have AH = AC.sin(x)
Then AH = 39.(2.sin(x).cos(x)) = 65.sin(x). That gives cos(x) = (39.2)/65 = 5/6. Using sin(x)^2 + cos(x)^2 = 1, we have then that cos(x)^2 = 1 - (25/36) = 11/36.
Finally sin(x) = sqrt(11)/6. Now we use the formula sin(3.x) = 3.sin(x) -4.sin(x)^3 to calculate sin(3.x)
sin(3.x) = 3.(sqrt(11)/6) - 4. (11.sqrt(11))/ 216 = (1/216). 108.sqrt(11) - 44.sqrt(11)) = (64/216).sqrt(11) = (8/27).sqrt(11)
Now the area of the triangle ABC is (1/2).AB.AC.sin(angle BAC) = (1/2). 65. 39. sin (180° -3.x) = (1/2).65.39. sin(3.x) = (1/2).65.39.((8/27).sqrt(11))
Finally the area of the triangle ABC is (3380/9).sqrt(11) when simplified. (about 1245.57)

[ОльгаСоломашенко-ьы]

Why find the degree measure of angles? It is enough to find sin(x), cos(x), sin(2x), cos(2x) using the sine theorem and basic trigonometric identities. Next, find the height drawn to the BC side and the length of the BC side.

[RobertHering-tqbn]

Hello, I did it as suggested as "Home Work". I started with the triangle ABC and angles x, 2x and 180°-3x. I added a line from C to AB, crossing AB in a point D by deviding the angle at C in two equal ones with value x. Then angle BDC is 180°-2x and angle ADC is 2x. As a result triangles ABC and ACD are similar, because all angles are the same. In addition we have BD = CD because the angles CBD and BCD are both x. I named this length d. BC is a, AC is b and AB is c. In addition AD is named e. Now we have...

b:c = e:b <==> e = b²/c = 39²/65 = 117/5

d = BD = c - e = 65 - 117/5 = (325-117)/5 = 208/5

a:CD = c:b <==> a = d*(c/b) = (208/5)*(65/39) = (208/5)*(5/3) = 208/3

Now we can use Heron's Formula to find the area of triangle ABC.

s = (a + b + c)/2 = (208/3 + 39 + 65)/2 = (208 + 117 + 195)/6 = 520/6 = 260/3

Area(ABC) = Sqrt(s(s-a)(s-b)(s-c)) = = =
(4*5*13²)/(3*3)*Sqrt(11) = (3380/9)*Sqrt(11)

And this value is about 1245.5767 which is not 100% what is shown in the video but may be a result of the different way of calculating.

[jimlocke]

PreMath usually provides exact answers, with a decimal approximation calculated at the end. I worked out an exact answer as follows: At about 5:30, PreMath finds that cos(x) = 65/78, which simplifies to 5/6. We can now construct a right triangle, off to the side, with angle x, the side adjacent to x has length 5, the hypotenuse, 6, and opposite side, by Pythagorean theorem, √(11). From that triangle, sin(x) = (√(11))/6 and tan(x) = (√(11))/5. Applying the tangent double angle formula, tan(2x) = 2tan(x)/(1 - tan²(x)) = (2)(√(11))/5)/(1 - (√(11))/5)²) = ((2√(11)/5)/(1 - ((11)/(25))) = ((2√(11)/5)/((14)/(25)) = 5(√(11))/7. So, we can construct a second right triangle, also off to the side, with angle 2x, the side opposite = 5√(11), the adjacent side 7 and, by Pythagorean theorem, hypotenuse 18. From that triangle, sin(2x) = (5√(11))/18 and cos(2x) = 7/18. We drop a perpendicular from A to BC, labelling the intersection as point D. We note that BD = (65)cos(x) = (65)(5/6) = (325)/6, and CD = (39)cos(2x) = (39)(7/18) = (273)/(18) = (91)/6 and combined length BC = (325)/6 + (91)/6 = (416)/6. Length AD = (65)sin(x) = (65)(√(11))/6. Let BC be the base and AD the height, then area ΔABC = (1/2)bh = (1/2)((416)/6)(((65)√(11))/6) = (27040)(√(11))/72 = (13520)(√(11))/36. The decimal equivalent, 1245.58, is probably within roundoff error of PreMath's result.

[tombufford]

At a quick glance: Using the Trigonometric relation, sin(2x) =2*sin(x)*cos(x), Law of sines and area of triangle formula, Ratio 39/sin(x) = 65/sin(2*x) = BC/ (sin (a) where a = 180 - 3*x. area = 0.5 * BC * h where h is height. sin(2 * x) = 2*sin(x) * cos(x). Then 65 / 2*sin(x) *cos(x) = 39 /sin(x). Then Cos(x) = 65/(2*39) . Then x=33.6 and a = 180 - 3 * x = 79.2. Then BC=65 * sin(79.2)/sin(67.1). Then BC = 69.3. Then h = 65 * sin(33.6) = 36 and area = 18 * 69.3 = 1247.4 area units.

[LuisdeBritoCamacho]

Using a Geometric Calculator I can realize that triangle [ABC] is is an isosceles triangle; BC = AB = 65 lin un. a = 65 ; b = 39 ; c = 65.
Then angle ABC = xº ; angle BCA = angle BAC = 2xº. 5xº = 180º.
The height of point A, distance to line BC ~ 37, 204 lin un
Area = (65 * 37, 204) / 2 = 1.209, 13 sq un
Answer:
The Area of Triangle ABC is approx. equal to 1.209, 13 Square Units.

[KipIngram]

This is easy. Note that from right triangle rules we have 65*sin(x) = 39*sin(2x). Use the sin(2x) formula to rewrite that as 65*sin(x) = 39*2*sin(x)*cos(x). Divide out sin(x) and we're left with cos(x) = 65/78, or cos(x) = 5/6. Now we know x. Then it's easy to get the base of the triangle as 65*cos(x) + 39*cos(2x), and the height as 65*sin(x). Then the area is half the product of base and height.

[handy]

Cannot you just have a line from A to point D on BC so that AD = AC = 39 and ADC angle = 2x Angle CAD = 180-4x so angle DAB = x and iscosceles triangle so BD = 39 and then use cosine rule to find angle x and then find 180-3x and use sine rule?

[murdock]

This is amazing, many thanks, Sir!
∆ ABC → AB = 65; AC = 39; BC = BM + CM → sin⁡(BMA) = 1; ABC = δ; BCA = 2δ; area ∆ ABC = ?
sin⁡(δ)/39 = sin⁡(2δ)/65 = 2sin⁡(δ)cos⁡(δ)/65 → cos⁡(δ) = 5/6 →
sin⁡(δ) = √(1 - cos^2 (δ)) = √11/6 → sin⁡(2δ) = 2sin⁡(δ)cos⁡(δ) = 5√11/18 →
cos⁡(2δ )= √(1 - sin^2(2δ)) = 7/18 = CM/39 → CM = 91/6
cos⁡(δ) = 5/6 = BM/65 → BM = 325/6 →
BC = BM + CM = 208/3 → area ∆ ABC = (1/2)sin⁡(δ)(65)(208/3) = (5√11)(26/3)^2

[unknownidentity]

Let's find the area of the triangle:
.
..
...
....


From the law of sines we can conclude:


AB/sin(∠ACB) = AC/sin(∠ABC)
AB/sin(2x) = AC/sin(x)
AB/AC = sin(2x)/sin(x)
AB/AC = 2*sin(x)*cos(x)/sin(x)
AB/AC = 2*cos(x)
⇒ cos(x) = AB/(2*AC) = 65/(2*39) = 5/6

The area of the triangle can now be calculated as follows:

A(ABC)
= (1/2)*AB*AC*sin(∠BAC)

sin(∠BAC)
= sin(180° − ∠ABC − ∠ACB)
= sin(180° − x − 2x)
= sin(180° − 3x)
= sin(3x)
= sin(2x)*cos(x) + cos(2x)*sin(x)
= 2*sin(x)*cos²(x) + [cos²(x) − sin²(x)]*sin(x)
= sin(x)*[2*cos²(x) + cos²(x) − sin²(x)]
= √[1 − cos²(x)]*[3*cos²(x) − 1 + cos²(x)]
= √[1 − cos²(x)]*[4*cos²(x) − 1]
= √(1 − 25/36)*(4*25/36 − 1)
= √(36/36 − 25/36)*(25/9 − 1)
= √(11/36)*(25/9 − 9/9)
= (√11/6)*(16/9)
= 8*√11/27

A(ABC) = (1/2)*39*65*8*√11/27 = 3380*√11/9 ≈ 1245.58

Best regards from Germany

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال H المسقط العمودي لAعلى (BC)
....
cos x=65/78
BH=325/6
AH=65(جذر11)/6
CH=91/6
BC=208/3
S=52×65×(11جذر)/9

[himo]

BD=DA=AC=39
AE=h DE=EC=y
h^2=65^2-(39+y)^2 h^2=39^2-y^2
4225-1521-78y-y^2=1521-y^2
78y=1183 6y=91 y=91/6

1521-8281/36=54756/36-8281/36
=46475/36
h=5√1859/6

=208/3

area of triangle ABC :
5√1859/6＊208/3＊1/2=260√1859/9

[implzeRX]

Sir Triangle [ABC] IS AN RIGHT ANGLE pgt....If we see 65²=39²+y²....by this we get
here we get 3rd side therefore area = ½xbase x about this sir??

[SkinnerRobot]

Many thanks for your excellent video, PreMath.

I noticed that you used a lowercase c in your review of the triange area formula, i.e. ½ab⋅sin c. I think it should be an uppercase C to distinguish it from the side c, which is opposite to the angle C. So, the formula should read ½ab⋅sin C.

[DB-lgsq]

شكرا لكم على المجهودات
القيمة المضبوطة للمساحة المطلوبة هي
9 /65×13×4(جذر 11)

[misterenter-izrz]

cos x=65/78, clearly, sin x=sqrt(1859)/78, height=65sin x=(65/78)sqrt(1859), base=65cos x+39cos 2x =65^2/78+39(2×(65/78)^2-1), therefore the area is

[joeschmo]

I dropped the altitude from A to the base, and equated sines (opposite/hypotenuse)...