# Sine and Cosine Rule - Worked Example 4

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.

## Case 1 - A is obtuse or a right angle.

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.

## Case 2 - A is acute and a is bigger than b

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°

## Case 3 - b is equal to a

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

## Case 4 - a is less than b

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-1B, 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