   Chapter 5.5, Problem 33E 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: ∆ A B C with m ∠ A = 45 ° , m ∠ B = 30 ° , and B C = 12 Find: A B (HINT: Use altitude C D ¯ from C to A B ¯ as an auxiliary line) To determine

To find:

AB of the ABC with mA=45°, mB=30° and BC=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)

Approach:

For a right triangle for which the measure of the interior angles 45°, 45°, and 90°, if ‘a’ is the length of measure of one of the leg; opposite to the angle 45°, then the length of the other two sides is given by

Length of the other leg =a

Length of the hypotenuse =a2

In general

Length of the legs are equal.

Length of the hypotenuse =2× (Length of one of the legs)

Calculation:

Given,

ABC with

mA=45°,

mB=30° and

BC=12.

Let us find the third angle C of the ABC.

By the property that the sum of all the three interior angle of a triangle is 180°.

A+B+C=180°

45°+30°+C=180°

75°+C=180°

C=105°

Draw the altitude CD¯ from C to AB¯, which divides the ABC into two right triangles ACD and right BCD and also divides the angle at the vertex C of the ABC into two parts.

One part of the angle C in the right ACD is 45° and the other part should be 60°, which lies in the right BCD such that the sum of all the three interior angles in both the right ACD and in the right BCD become 180°

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