computational <A= 3<B, BC=5, CA=3, AB=? (2008 Argentina OMA L2 p3)

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

computational <A= 3<B, BC=5, CA=3, AB=? (2008 Argentina OMA L2 p3)

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30650 posts

#1 h Jun 12, 2020, 9:15 PM

Y by

Let $ABC$ be a triangle such that $\angle A= 3\angle B$ . If $BC = 5$ and $CA = 3$ , calculate the measurement of side $AB$ .

This post has been edited 1 time. Last edited by parmenides51, Jun 12, 2020, 9:15 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

8219 posts

#2 h Jun 12, 2020, 10:20 PM • 1 Y

Y by Mango247

We know $\sin 3\theta=3\sin \theta-4\sin^3 \theta$
Define $\sin \angle B=x$
By LOS we know $\frac{x}{3}=\frac{3x-4x^3}{5}$
$5x=9x-12x^3$
$x=\frac{1}{\sqrt{3}}$

$\sin(180-4\theta)=\sin(4\theta)=4\sin \theta \cos\theta(1-2\sin^2\theta)$
So $\sin \angle C=\sin (180-4\angle B)=4x(1-2x^2)\sqrt{1-x^2}=\frac{4\sqrt{2}}{9}$

Using LOS once again, we get $AC=\frac{4\sqrt{6}}{3}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a