Double Trouble in Triangle

Geometry Level 4

Let A B C ABC be a triangle with A C = 28 , B C = 33 , A B C = 2 A C B AC = 28,BC = 33, \angle ABC = 2\angle ACB . Compute the length of the side A B AB .


The answer is 16.

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2 solutions

Ayush G Rai
Oct 9, 2016

the paper is very small can u magnify a little more

abhishek alva - 4 years, 8 months ago

Almost same as Mr. Ayush Rai .
U s i n g S i n l a w : c S i n C = 28 S i n 2 C = 33 S i n ( π 3 C ) . S i n ( π 3 C ) = S i n 3 C = S i n C ( 4 C o s 2 C 1 ) , a n d D i v i d i n g b y S i n C g i v e s , c = 28 2 C o s C = 33 4 C o s 2 C 1 . S o l v i n g t h e l a s t e q u a t i o n , 14 ( 4 C o s 2 C 1 ) 33 C o s C = 0 , a q u a d r a t i c i n C o s C , a n d s i n c e C < π 2 , C o s C > 0 , C o s C = 14 16 , c = A B = 28 2 14 16 = 16 Using~Sin~law:-\\ \dfrac c {SinC}=\dfrac {28}{Sin2C}=\dfrac {33}{Sin(\pi-3C)}.\\ Sin(\pi-3C)=Sin3C=SinC(4Cos^2C-1), ~~~~and~~~~Dividing~by~SinC~gives,\\ c~ =\dfrac {28}{2CosC}=\dfrac {33}{4Cos^2C-1}.\\ Solving~the~last~equation,~14*(4Cos^2C-1)-33CosC=0,\\ a~quadratic ~in ~CosC,~~and~ since~C<\frac \pi 2,~~CosC>0,\\ CosC=\dfrac{14}{16},\\ \implies~~c=AB=\dfrac {28}{2*\frac{14}{16}}=\Large~~\color{#D61F06}{16}

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