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.

Thank u all.

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.

Thank u all.

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TopNewestLet 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. – Clarence Chew · 3 years, 10 months ago

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Infact, If in an equilateral triangle, D is a point such that \(AD^2+BD^2=CD^2\), 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 \(AC=BC ACD=BCE DC=EC\) So triangle \(ACD\) and triangle \(BCE\) are congruent. So AD=BE. Thus \(BD^2+DE^2\) = \(BD^2+CD^2\) =\(AD^2\) =\(BE^2\) So \(BDE\) is 90 degrees, by pythagoras theorem. So \(BDC=BDE+CDE=150\). – Shourya Pandey · 3 years, 10 months ago

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– Shourya Pandey · 3 years, 10 months ago

Sorry, the 4th line in para 2 is \(AC=BC\) \(ACD\)=\(BCE\) \(DC=EC\)Log in to reply

my math is so weak i wanna command in math so how do my problem solve plz tell me – Daniyal Arain · 3 years, 10 months ago

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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. – Falensius Nango · 3 years, 10 months ago

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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 . – Rashmi Rastogi · 3 years, 10 months ago

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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) – Srinath Reddy · 3 years, 10 months ago

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150 degrees. – Shourya Pandey · 3 years, 10 months ago

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such a easy problem and u r hobbling over it!!!!!!!!!!!!!!!!!!!!!!!!!!!! – Sayan Chowdhury · 3 years, 10 months ago

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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... – Gabriel Singhal · 3 years, 10 months ago

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– Swarandeep Singh · 3 years, 10 months ago

if its so then AD=BD=CD, here we apply cosine rule ....but with diff approachLog in to reply