30-60-90

Geometry Level 2

The length of CP is 4. What is the length of AB?

3 2 2 3\sqrt{2}-2 2 3 2\sqrt{3} 3 + 1 \sqrt{3}+1 3 + 2 3 3+\sqrt{2}-\sqrt{3}

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Chuen-Wei Chen
Oct 6, 2016

BP = CP = 4; AB = BP*cos30

Viki Zeta
Oct 6, 2016

C B A = 60 tan ( C B A ) = tan ( 60 ) = 3 A C A B = 3 4 + A P A B = 3 4 + A P 3 = A B . . . ( i ) P B A = 30 tan ( 30 ) = 1 3 A P A B = 3 A B = 3 A P . . . ( i i ) (i) and (ii) 4 + A P 3 = 3 A P 4 + A P = 3 A P A P = 2 A B = 3 A P = 3 2 = 2 3 \angle CBA = 60 \\ \tan(\angle CBA) = \tan(60) = \sqrt[]{3} \\ \dfrac{AC}{AB} = \sqrt[]{3} \\ \dfrac{4+AP}{AB} = \sqrt[]{3} \\ \dfrac{4+AP}{\sqrt[]{3}} = AB ...(i)\\ \angle PBA = 30 \\ \tan(30) = \dfrac{1}{\sqrt[]{3}} \\ \dfrac{AP}{AB} = \sqrt[]{3} \\ AB = \sqrt[]{3}AP ... (ii) \\ \text{(i) and (ii)} \\ \implies \dfrac{4+AP}{\sqrt[]{3}} = \sqrt[]{3}AP\\ 4 + AP = 3AP \\ AP = 2\\ \boxed{\therefore AB = \sqrt[]{3}AP = \sqrt[]{3}2 = 2\sqrt[]{3}}

Maria Kozlowska
Oct 6, 2016

The image shows half of an equilateral triangle with point P P its incenter. That makes A P = 4 / 2 = 2 AP=4/2=2 , A B = 2 / tan ( 30 ) = 2 3 AB=2/\tan(30)=2\sqrt{3} .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...