   Chapter 5.5, Problem 32E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

In Exercises 27 to 33, give both exact solutions and approximate solutions to two decimal places. Given: Right ∆ A B C with m ∠ C = 90 ° and m ∠ B A C = 60 ° ; points D on B C ¯ ; A D ⃑ bisects ∠ B A C and A C ¯ = 2 3 Find: B D To determine

To find:

BD in a right ABC such that mC=90° and mBAC=60°; point D on BC¯;AD

Bisects BAC and AC=23.

Explanation

Approach:

For a right triangle, for which the measure of the interior angles 30°, 60°, and 90°; if ‘a’ is the length of measure of the shorter leg; opposite to the angle 30°, then the length of the other two sides is given by

Length of the longer leg (opposite to 60°) =a3

Length of the hypotenuse (opposite to 90°)=2a.

In general

Length of the longer leg =3× (Length of the shorter leg)

Length of the hypotenuse =2× (Length of the shorter leg)

Calculation:

Given,

A right ABC,

mC=90° and mBAC=60°; point D on BC¯;AD

Bisects BAC and AC=23.

Since, AD bisects A of the ABC, we have

=12(60°)

First consider the right ABC to find AB and BC.

The above should be of the type 30°-60°-90° as such mA=60°, mC=90°.

Thus,

mB=30°

30°-60°-90° theorem

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