Can you find the length CP? | (Fun Geometry) |

[@aprendendomatematica700]

The triangles ∆Q'PA and ∆Q'QR' are congruent by a-s-a then QQ'+Q'P=16 , 2QQ'=16 QQ'=8 and it is not dificult to see that the triangle ∆Q'QC is equitateral, so the segment QC=8 and PQC=60 Step 2 Now a²+b²-2abcos(x)=c² Given QP=16 PC=y QC=8 PQC=60 However, 16²+8²-2×8×16×cos(60)=y² y²=16²+8²-8×16 y²=8(8-16)+16² y²=8(8-16+4×8) y²=8(-8+4×8) y²=8×8(-1+4) y²=8²×3 y=8√3

[@aprendendomatematica700]

Elementary

[@MathandEngineering]

Wow the method is very good and also creative, I could have solved it using this method, I'll make sure I try this approach of yours on similar Questions, I am sure it will produce a precise answer like this

[@aprendendomatematica700]

@@MathandEngineering thanks, bring problems from Jee advence, please

[@MathandEngineering]

Ok I'll go look for the Question, and please, if there is any Questions you want me to solve, you can also put it in the comment section, I'll solve it, thanks

[@aprendendomatematica700]

​@@MathandEngineeringIn triangle ABC, show that sin( A /2)≤ a/b+c Try this .