# Trig! Exam tomorrow!

• May 26th 2010, 11:24 AM
Trig! Exam tomorrow!
Hello,

I am here reviewing for my exam tmw.. this question I can't figure out - pease help.

Two triangular building lots, ABC and ACD, have side AC in common.
(in attachment)

a) Determine an exact expression for the side length of side AC
b) Determine an exact expression for the length of side AB

I get confused from the beginning because I don't know which trig ratio to solve with? (Sin/Tan or Cos?)
• May 26th 2010, 11:55 AM
masters
Quote:

Hello,

I am here reviewing for my exam tmw.. this question I can't figure out - pease help.

Two triangular building lots, ABC and ACD, have side AC in common.
(in attachment)

a) Determine an exact expression for the side length of side AC
b) Determine an exact expression for the length of side AB

I get confused from the beginning because I don't know which trig ratio to solve with? (Sin/Tan or Cos?)

These angles are special angles.

$\displaystyle \frac{\pi}{4}=45^{\circ}\:\:and\:\:\frac{\pi}{6}=3 0^{\circ}$

Use your rules for a 45-45-90 triangle and then use your
rules for a 30-60-90 triangle.

First,

$\displaystyle 60=AC\sqrt{2}$

$\displaystyle AC=\frac{60}{\sqrt{2}}$

Rationalizing the denominator, we get

$\displaystyle AC=\frac{60}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{60\sqrt{2}}{2}=30\ sqrt{2}$

This value is the hypotenuse of the 30-60-90 triangle,
so we have CB as the short leg and it's equal to one-half the hypotenuse...

$\displaystyle CB=15\sqrt{2}$

And...AB is the long leg which is the short leg times the square root of 3.

$\displaystyle AB=15\sqrt{2}\cdot \sqrt{3}=15\sqrt{6}$