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).
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.
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
Very good method. Excellent
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Good evining Sir
Thanks for your efforts
That’s very nice
With my glades
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Fine.
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For me it was a teaching moment.
Following the method and overcoming the confusion of the figure being not to scale (😂)
Great problem!
So cool! The Pythagorean Theorem is a Special Case of the Law of Cosines. 🙂 Had a lot of fun solving. 😉
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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).
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there are total 3 methods to solve it wow
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.
The law of cosines rules.
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1/2 × root 3 / 2 ×7 ×37?
111√3/4
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Sir dint understood. Pls explain using simple method
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
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Exactly this is how I solved it 😄
@@ramanan36what a coincidence! You must be quite a trigonometry lover as me.