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Answers To The Exercises: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14
Lesson 2
 


2.1.
See footnote on the first page of Lesson 2

2.2.
The main elements are three sides and three angles.

2.3.
It means to find main elements (sides and angles), given only some of them.

2.4.
To solve a triangle, three main elements are needed, and among them, at least one side must be given. Otherwise, the problem will have infinitely many solutions.

2.5.
There are four problems for solving triangles:
1) One side and two angles are given.
2) Two sides and an angle in between are given.
3) Three sides are given.
4) Two sides and an angle opposite to one of the sides are given.

We obtained exactly four problems because they represent all possible combinations of three main elements, and at least one of them is a side.

2.6.
The fourth problem (see the answer to exercise 2.5) can have two solutions.

2.7.
1) One side and two angles are given: both angles are obtuse angles.
2) Two sides and an angle in between are given. The angle is a straight one.
3) Three sides are given: the sum of two sides is less than or equal to the third side.
4) Two sides and an angle opposite to one of the sides are given: the angle is the right one, the opposite side is less then another side. Such a triangle does not exist because hypotenuse can not be less than a leg.

2.8.
The solutions can be seen from the following figures:

2.9.
The problem may have no solution or one solution.
1) The problem has no solution if and only if b a.
2) The problem has one solution if and only if b > a.

2.10.

2.11.

2.12.
The problem has two solutions


2.13.

BC2 = AC2 + AB2 – 2·AC·AB·cos 60°= 64 + 25 – 2·8·5·0.5 = 49.   BC = 7.

BC/sin 60° = AC/sin B;   7/0.87 = 8/sinB;   sinB= 8·0.87/7= 0.99.    B = 82°.

2.14.

2.15.

 

2.16.

2.17.

Draw the height BD. Then BD = c·sin A
From here, Area = 1/2·b·BD =1/2 bc sin A

 

 

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