Question

In a triangle ABC, the measure of angle A is fifteen less than twice the measure of angle B. The measure of angle C equals the sum of the measures of angle A and angle B. What is the measure of angle B?

Answer 1

If the reference to "fifteen" means `15^@`
then `B=35^@`.

If the refrence to "fifteen" means `15` radians
then `B=pi/6+5` radians.

Explanation:

We are told explicitly that
[1]`color(white)("XXXX")A+15 = 2B`
[2]`color(white)("XXXX")C=A+B`
and, implicitly (assuming the triangle lies in a Cartesian plane)
[3]`color(white)("XXXX")A+B+C=180 [pi]`
`color(white)("XXXX")`note: I will use degrees as the default assumption but place radian measures in square brackets when it differs.

Arranging these into standard form:
[4]`color(white)("XXXX")A-2B=-15`
[5]`color(white)("XXXX")A+B-C=0`
[6]`color(white)("XXXX")A+B+C=180color(white)("XXX") [pi]`

Adding [5] and [6]
[7]`color(white)("XXXX")2A+2B=180color(white)("XXX") [pi]`
Multiplying [4] by 2
[8]`color(white)("XXXX")2A-4B=-30`
Subtracting [8] from [7]
[9]`color(white)("XXXX")6B = 210color(white)("XXX") [pi+30]`
Dividing by 6
[10]`color(white)("XXXX")B=35color(white)("XXX") [pi/6+5]`

Answer 2

`B=35^@`

Explanation:

First of all, let's translate this problem into mathematical language. I suppose that we are working with degree measures of angles (so that writing "fifteen" you meant "fifteen degrees").
`A=2B-15^@`
`C=A+B`
We also have to "decode" another crucial information: these are angles in a triangle, so their sum equals `180^@`:
`A+B+C=180^@`

We want to find the measure of the angle `B` and we have 3 equations and 3 unknowns.
The measure of the angle `C` is very easy to find, so we compute it to reduce the number of variables involved: since `A+B=C`, we take the third equation and substitute the second one
`180^@=A+B+C=(A+B)+C=C+C=2C`
So dividing both sides by `2`, we get that `C=90^@` is a right angle and the second equation turns to `A+B=90^@`, meaning that `A=90^@-B`.

Now we substitute this last result in the first equation:
`2B-15^@=A=90^@-B`
and we get an equation where `B` is the only unknown
`2B-15^@=90^@-B`
Let's solve this:
`B+(2B-15^@)=(90^@-B)+B` (sum `B` on both sides)
`3B-15^@=90^@`
`15^@+(3B-15^@)=(90^@)+15^@` (sum `15^@` on both sides)
`3B=105^@`
`(3B)/3=(105^@)/3` (divide both sides by `3`)
`B=35^@`

Note: The only angle left is `A` and the computation of `A` can be done using one of the 3 former equations. We get that `A=55^@`.