   Chapter 11.CR, Problem 14CR ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
1 views

# In Exercises 13 to 17, use the Law of S i n e s or the Law of C o s i n e s to solve each problem. Angle measures should be found to the nearest degree; distances should be found to the nearest tenth of a unit.Two sides of a parallelogram are 50 cm and 70 cm long. Find the length of the shorter diagonal if a larger angle of the parallelogram measures 105 ° .

To determine

To find:

The length of the shorter diagonal of a parallelogram.

Explanation

Procedure used:

In any triangle ABC such that

AB=c,

BC=a,

CA=b,

mA=α,

mB=β, and

mC=γ.

The Law of cosines is given by

a2=b2+c2-2bccosα

b2=c2+a2-2cacosβ

c2=a2+b2-2abcosγ

The Law of sines is given by

asinα=bsinβ=csinγ

Calculation:

Given:

Length of the sides of the parallelogram are 50 cm and 70 cm.

The larger angle =105°

Let ABCD be the parallelogram with diagonals AC and BD, which intersect at ‘O’.

From the above figure we have

CD=AB=50 cm

The diagonal BD is shorter than the diagonal AC.

To find the length of the shorter diagonal BD, it is necessary to find the angle mBAD.

By the property of parallelogram, consecutive angles of any parallelogram is supplementary

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 