Okay, these problems are not really random, they’re from the same section of a textbook:

This book is called Geometry Revisited (by Coxeter and Greitzer). It’s often recommended as a olympiad level geometry textbook. I would recommend this book too (although I haven’t gone through the entire book yet).

Section 1.1 of this book proves a theorem referred to as the **Extended Law of Sines**:

where is the circumradius of triangle .

The extended Law of Sines is often given in a truncated form, without mention to the circumradius.

After the proof the book gives four exercises, which I shall rephrase and solve here.

### Problem 1

Show that in any triangle ,

From this, deduce the ‘addition formula’ for sines:

#### Solution

From the Law of Cosines:

Adding the two:

as desired.

We now prove the addition formula. From the law of sines:

Similarly:

Substituting into our previous equation:

Since is a triangle, . Thus, .

Note that . Thus,

as desired.

### Problem 2

Show that in any triangle ,

#### Solution

Using the law of sines again:

Similarly,

Adding the three equations:

as desired.

### Problem 3

Show that in any triangle ,

(although the book uses , I’ve accustomed to writing on my blog to denote area of )

#### Solution

Let *h* be the height of (from *A* perpendicular to *BC*), so that :

Using trigonometry,

so we can write the area as

From the extended Law of Sines,

Now we can rewrite the area again:

as desired.

### Problem 4

For a triangle , let *p *be the radius of a circle going through *A *and tangent to *BC *at *B*. Let *q *be the radius of a circle going through *A* and tangent to *BC* at *C*.

Show that:

#### Solution

Let M be the center of the circle with radius *p*, and *N *be the center of the circle with radius *q*. Also *K *and *L* are two points on circles *M* and *N* respectively:

Using the law of sines on circle *N* and *M*, we have:

Because *BC* is tangent to the two circles, and .

Thus,

Now applying the Law of Sines again in :

Substituting these values back into the previous equation:

By multiplication,

as desired.