   Chapter 5.5, Problem 27E 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: In ∆ A B C , A D ⃑ bisects ∠ B A C m ∠ B = 30 ° and A B = 12 Find: D C and D B To determine

To find:

DC and DB such that in ABCD, AD bisects BAC, mB=30° and AB=12.

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,

The ABC with bisector of the angle BAC.

mB=30°

AB=12

Since, one of the acute angle mC of the right triangle ABC is 30°, then the ofter acute angle mA should be 60°

Thus, the ABC is of the type 30°-60°-90° triangle.

Since, the interior angle A of the triangle ABC is bisected by the ray AD, we have

Thus,

Now, consider the right triangle ABC

Thus, ABC is of the type 30°-60°-90° triangle.

AB= Hypotenuse =12 units.

CA= Shorter leg,

BC= Longer leg.

30°-60°-90° Theorem.

In a right triangle whose angle measure 30°, 60°, and 90°, the hypotenuse has a length equal to twice the length of the shorter leg, and the longer leg is the product of 3 and the length of the shorter leg.

The Hypotenuse =2× (Length of shorter leg)

AB=2×(AC)

12=2×(AC)

Divide; by 2 on both sides,

122=2×AC2

AC=6 units

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