I have a basic geometry problem. It is given equilateral triangle ABC and a point D inside the triangle. The length of AD = 8 cm, BD = 15 cm, and CD = 17 cm. Determine the angle ADB and the length of sides of triangle ABC.
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Let E be a point where Triangle BDC=Triangle BEA. Then you have another small equilateral triangle (BDE) and a right triangle, and you can finish the problem.
Infact, If in an equilateral triangle, D is a point such that AD2+BD2=CD2, then angle ADB=150 degrees.
The proof involves construction.
Draw an equilateral triangle with one of the sides as DC. There are two such triangles possible, draw the one with the third vertex, say E, nearer to B. Join DE and BE. So triangle DCE is equilateral.
Now note that angle CDE=60.
also , DCE=BCA=60, so BCE=ACD.
Thus, we have AC=BCACD=BCEDC=EC
So triangle ACD and triangle BCE are congruent.
So AD=BE.
Thus BD2+DE2
= BD2+CD2
=AD2
=BE2
So BDE is 90 degrees, by pythagoras theorem.
So BDC=BDE+CDE=150.
Three circles of radius 3cm , 5cm and 7cm . touches each other externally . find the radius of the circle passing through the centres of the three circles .
You my take the point D to be the orthocentre , incentre,etc. For equilateral triangle they are a single point. Then u can easily calculate the angle by a little bit of Cosine Rule. But for a subjective solution, u need to think of a different approach. My method works only for calculating the answer...
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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Let E be a point where Triangle BDC=Triangle BEA. Then you have another small equilateral triangle (BDE) and a right triangle, and you can finish the problem.
Infact, If in an equilateral triangle, D is a point such that AD2+BD2=CD2, then angle ADB=150 degrees. The proof involves construction.
Draw an equilateral triangle with one of the sides as DC. There are two such triangles possible, draw the one with the third vertex, say E, nearer to B. Join DE and BE. So triangle DCE is equilateral. Now note that angle CDE=60. also , DCE=BCA=60, so BCE=ACD. Thus, we have AC=BCACD=BCEDC=EC So triangle ACD and triangle BCE are congruent. So AD=BE. Thus BD2+DE2 = BD2+CD2 =AD2 =BE2 So BDE is 90 degrees, by pythagoras theorem. So BDC=BDE+CDE=150.
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Sorry, the 4th line in para 2 is AC=BC ACD=BCE DC=EC
my math is so weak i wanna command in math so how do my problem solve plz tell me
150 degrees.
Three circles of radius 3cm , 5cm and 7cm . touches each other externally . find the radius of the circle passing through the centres of the three circles .
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The radius of the circle passing through the three centres is the radius of the circumcircle of the triangle formed by the three points.
The triangle has sides 8 ,10,12. We can find the radius of the circumcircle with the formula (abc)/(4 x area of the triangle)
I have got an answer of 16/(7^0.5)
I found that this problem is not simple as i think in the solution. Thanks for helping me out and give me the elegant solution.
such a easy problem and u r hobbling over it!!!!!!!!!!!!!!!!!!!!!!!!!!!!
You my take the point D to be the orthocentre , incentre,etc. For equilateral triangle they are a single point. Then u can easily calculate the angle by a little bit of Cosine Rule. But for a subjective solution, u need to think of a different approach. My method works only for calculating the answer...
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if its so then AD=BD=CD, here we apply cosine rule ....but with diff approach