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?

2 Answers
Oct 31, 2015

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

If the refrence to "fifteen" means 1515 radians
then B=pi/6+5B=π6+5 radians.

Explanation:

We are told explicitly that
[1]color(white)("XXXX")A+15 = 2BXXXXA+15=2B
[2]color(white)("XXXX")C=A+BXXXXC=A+B
and, implicitly (assuming the triangle lies in a Cartesian plane)
[3]color(white)("XXXX")A+B+C=180 [pi]XXXXA+B+C=180[π]
color(white)("XXXX")XXXXnote: 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=-15XXXXA2B=15
[5]color(white)("XXXX")A+B-C=0XXXXA+BC=0
[6]color(white)("XXXX")A+B+C=180color(white)("XXX") [pi]XXXXA+B+C=180XXX[π]

Adding [5] and [6]
[7]color(white)("XXXX")2A+2B=180color(white)("XXX") [pi]XXXX2A+2B=180XXX[π]
Multiplying [4] by 2
[8]color(white)("XXXX")2A-4B=-30XXXX2A4B=30
Subtracting [8] from [7]
[9]color(white)("XXXX")6B = 210color(white)("XXX") [pi+30]XXXX6B=210XXX[π+30]
Dividing by 6
[10]color(white)("XXXX")B=35color(white)("XXX") [pi/6+5]XXXXB=35XXX[π6+5]

B=35^@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^@A=2B15
C=A+BC=A+B
We also have to "decode" another crucial information: these are angles in a triangle, so their sum equals 180^@180:
A+B+C=180^@A+B+C=180

We want to find the measure of the angle BB and we have 3 equations and 3 unknowns.
The measure of the angle CC is very easy to find, so we compute it to reduce the number of variables involved: since A+B=CA+B=C, we take the third equation and substitute the second one
180^@=A+B+C=(A+B)+C=C+C=2C180=A+B+C=(A+B)+C=C+C=2C
So dividing both sides by 22, we get that C=90^@C=90 is a right angle and the second equation turns to A+B=90^@A+B=90, meaning that A=90^@-BA=90B.

Now we substitute this last result in the first equation:
2B-15^@=A=90^@-B2B15=A=90B
and we get an equation where BB is the only unknown
2B-15^@=90^@-B2B15=90B
Let's solve this:
B+(2B-15^@)=(90^@-B)+BB+(2B15)=(90B)+B (sum BB on both sides)
3B-15^@=90^@3B15=90
15^@+(3B-15^@)=(90^@)+15^@15+(3B15)=(90)+15 (sum 15^@15 on both sides)
3B=105^@3B=105
(3B)/3=(105^@)/33B3=1053 (divide both sides by 33)
B=35^@B=35

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