Here we are given two sides and the angle opposite one of them, that is A, a and b

If we do not really understand what we are doing when we are solving a triangle like this we cannot get full marks in an exam. To understand why, it is helpful if we think about how we can *construct* (draw using ruler, protractor and compasses) triangle ABC when we are given A, a and b.

First we draw the horizontal line AC of length b. Then we draw the (long) line AE at angle A to AC. Finally we draw an arc of a circle, radius a and centre C. B is where this arc cuts the line AE.

There is only one triangle possible, but of course a must be bigger than b.

If A is obtuse or a right angle then we *know* that B and C must both be acute, so we can use the Sine Rule to find B and C without any problems.

Your calculator gives 0.4182531898 and this is stored in ANS, but you do not need to write it down. So B = sin^{-1} (ANS) Your calculator gives 24.72435309 and this is stored in ANS, but you will write down 25°

So C = 180 - 124 - ANS. Your calculator gives 31.27564691 and this is stored in ANS but you will write down 31°

Your calculator gives 69.50988119 but you will write down 70 mm.

So B = 25°, C = 31° and c = 70 mm.Only one triangle is possible.

If A is acute and a is bigger than b then we know that A is bigger than B so B must also be acute, and once we know B then we know whether C is acute or obtuse. So now we find B.

So c = 154 mm, B = 31° and C = 113°

If a = b then A = B and they must therefore both be acute. Only one triangle is possible.

So B = 74°, C = 32° and c = 50 mm

The green line is the perpendicular from C onto AE. So b is the hypotenuse of a right angled triangle, and g is opposite A in this triangle.

We can see that if a is less than b but is bigger than b × sin A then *two* triangles are possible, one with an acute angle at B and one with an obtuse angle at B, while if a is less than b × sin A then no triangle is possible: if we try to solve it we shall find that the value of a sine is greater than one and our calculator will give a maths error. (If a is exactly equal to b × sin A then B is a right angle but for reasons described in Appendix 3, for *Advanced Readers*, this is only possible if A is 30° and b is twice a.)

So in an exam, unless we have other information which will help us to decide whether B is obtuse or acute, to get full marks we must show that we understand that two different solutions are possible.

So B = sin^{-1} ANS. This has two values (between 0° and 180°), one acute and one obtuse, and together they add up to 180°, but your calculator gives only the acute value. It says 66.42631336 and this is stored in ANS but you will write down 66°

The obtuse value is 180 - ANS and is 114° to the nearest degree. You will lose marks in an exam if you do not write down both values. Of course there are now two values for c and two values for C depending upon which value of B you use, but you will not usually be required to find both sets of values - but read the Question with more than ordinary care!

We are now going to find c and C using the acute value of B.

Use the calculation history facility of your calculator to return to find the acute value of sin^{-1}B, then C = 180 - 36 - ANS. Your calculator says 77.57368664 and this is stored in ANS but you will write down 78°.

Although B and C each have two values, if B is obtuse C *must* be acute, but if B is acute C is not *necessarily* obtuse.

Your calculator says 98.02532412 but you will write down 98 mm.

So B = 66°, C = 78° and c = 98 mm.

If we solve the triangle in the same way, but using the *obtuse* value of B we shall find that, to the nearest degree and millimetre, B = 114°, C = 30° and c = 51 mm.

© Barry Gray April 2016