Can you find area of the triangle ABC? | (Law of Cosines) |
Learn how to find the area of the triangle ABC. Important Geometry skills are also explained: Law of Cosines; similar triangles; exterior angle theorem. Step-by-step tutorial by PreMath.com
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Пікірлер: 37
Very good method. Excellent
@PreMath
26 күн бұрын
Many many thanks❤️🙏
Excellent method for finding x, but once you have x, at 9 minutes, the height of your triangle is xsin60 = 3rt3/2, so area is 1/2x37x3rt3/2 = 111rt3/4.
@jimlocke9320
26 күн бұрын
Yes, drop a perpendicular from C to AB and label the intersection as point E. Triangle CDE is a special 30-60-90 right triangle with hypotenuse CD = x = 3. Length CE = (3√3)/2 and is the height of ABC, if AB = 37 is treated as the base. Area = (1/2)(37)((3√3)/2) = (111√3)/4, as PreMath also found.
@PreMath
26 күн бұрын
Excellent! Thanks for the feedback ❤️
@allanflippin2453
26 күн бұрын
Yes, I also did it this way. Although I had no clue for finding "x", things got much simpler afterward.
Draw a line from C perpendicular to AD intersecting it at E. Then /_CED = 90 degrees, /_CDE = 60 degrees and /_ECD = 30 degrees. If CE = h, then ED = h/sqrt(3) from properties of a 30-60-90 degree triangle. Let /_CBE = alpha, then /_BCE = (90 - alpha) and /_ACE = 150 -(90- alpha) = 60 + alpha and /_CAE = 90 - (60 + alpha) = 30 - alpha (sum of angles of triangle CAE is 180 degrees). Next we use the sine rule on triangle CAE as follows: Sin(60 + alpha)/AE = Sin(30 - alpha)/CE or Sin(60 + alpha)/(7 - h/sqrt(3)) = Sin(30 - alpha)/h. ----(1) Then we apply the sine rule to triangle CBE as follows Sin(alpha)/CE = Sin(90 - alpha)/EB or Sin(alpha)/h = Sin(90 -alpha)/(30+h/sqrt(3)) ---(2) Next we expand the Sin angles and substitute the values for SIne and Cosine of angles 30, 60 and 90 degrees and end up with two equations for Tan(alpha) which we combine to get a quadratic equation for h: [h^2 + (201/(2*sqrt(3))*h - 315/2 = 0] which can be solved to give h = 2.59808. The area of triangle ABC is (1/2)*AB*CE = (1/2)*37*2.59808 = 48.064 (which is the same as (111*sqrt(3))/4).
@PreMath
26 күн бұрын
Excellent! Thanks for sharing ❤️
Very good
@PreMath
26 күн бұрын
Glad to hear that! Thanks for the feedback ❤️
I solved this by drawing a segment CE vertical to AD and found that ED = 1.5. I used the formula tan(A+B) = (tan A + tan B) / (1 - tan A * tan B), knowing that A + B = 30.
Good evining Sir Thanks for your efforts That’s very nice With my glades ❤❤❤❤❤
@PreMath
26 күн бұрын
Always welcome🌹 You are very welcome! Thanks for the feedback ❤️
Thank you!
@PreMath
26 күн бұрын
You bet! Thanks for the feedback ❤️
For me it was a teaching moment. Following the method and overcoming the confusion of the figure being not to scale (😂) Great problem!
Fine.
@PreMath
26 күн бұрын
Thanks for the feedback ❤️
So cool! The Pythagorean Theorem is a Special Case of the Law of Cosines. 🙂 Had a lot of fun solving. 😉
@PreMath
26 күн бұрын
So cool! Thanks for the feedback ❤️
there are total 3 methods to solve it wow
The law of cosines rules.
@PreMath
26 күн бұрын
Very true! Glad to hear that! Thanks for the feedback ❤️
I have an alternative way to find x=3 without having to construct any lines using trigonometry only. Here is my version. Let Angle CAB=a then angle ACD=120-a, angle ABC=30-a and angle BCD=30+a. Law of sine for triangle ACD, sin a/ x=sin(120-a)/7. ……(1) Similarly for triangle BCD, sin(30-a)/x=sin(30+a)/30. …..(2) Dividing (1) by (2) sin a/sin(30-a)=sin(120-a)/sin(30+a)*30/7 2*sin a*sin(30+a)=2*sin(30-a)*sin(120-a)*30/7 -cos(2*a+30)+cos 30=(-cos(150-2*a)+cos 90)*30/7 -cos(2*a+30)+sqrt(3)/2=(-cos(180-30-2*a)*30/7= -cos(180-(30+2*a))*30/7=cos(30+2*a)*30/7 cos(2*a+30)=7/74*sqrt(3) from which ,we get sin(2*a+30)=73/74 cos(2*a)=cos((2*a+30)-30)=cos(2*a+30)*cos 30+sin(2*a+30)*sin 30=47/74 cos(2*a)=2*cos(a)^2-1=47/74 from which tan(a)=3/11*sqrt(3) or cot(a)=11/(3*sqrt(3)) From (1) x=7*sin(a)/sin(120-a)=7*sin(a)/(sin(120)*cos(a)-cos 120*sin(a)=7/(sqrt(3)/2*cot(a)+1/2)=3
@PreMath
26 күн бұрын
Awesome! Thanks for sharing ❤️
@thinker821
25 күн бұрын
Exactly this is how I solved it 😄
@xualain3129
25 күн бұрын
@@thinker821what a coincidence! You must be quite a trigonometry lover as me.
1/2 × root 3 / 2 ×7 ×37?
111√3/4
@PreMath
26 күн бұрын
Excellent! Thanks for sharing ❤️
Cosine rule?no idea at all.😅
@Mediterranean81
26 күн бұрын
A more general formula of the Pythagorean theorem
@misterenter-iz7rz
26 күн бұрын
@Mediterranean81 I see. But no idea how to apply to solve this puzzle. 😕
@Mediterranean81
26 күн бұрын
@@misterenter-iz7rz calculate the sides using cosine rule then use 1/2absinc to find area