Step by step
The question
Question: find the area of the triangle with sides 5 and 12, and the angle 70° between them.
1 Find the height
The area needs a height.
Draw it from the top corner straight down to the base, at a square corner: h.
The small right triangle on the left has the side 12 as its longest side, and h across from the 70° angle.
SOH: sin is the opposite over the hypotenuse.
sin 70° = h/12.
To remove the 12 at the bottom, we multiply by 12.
We do it on both sides.
Why both sides?
The = sign means both sides are the same amount.
If we change only one side, they are not the same any more.
12/12 is 1.
So just h is left.
So h = 12 × sin 70°.
2 Write the rule
The area of a triangle is a half times the base times the height.
The base is 5, the height is 12 × sin 70°.
So the rule is ½ × a × b × sin C: two sides and the sine of the angle between them.
Careful!
The angle must be the one between the two sides we use.
Another angle gives a wrong height.
3 Put the numbers in
Now put the numbers in: the two sides 5 and 12, and sin 70°.
4 Work it out
5 times 12 is 60.
Half of 60 is 30.
The calculator gives sin 70°: about 0.9397.
Put the number in.
That is about 28.191.
To the nearest tenth, the area is about 28.2 square units.
5 Check
Let's check another way: use the side 12 as the base.
Its height is 5 times sin 70°: about 4.7.
A half times 12 times that is about 28.2.
The same area!
So it is right.
Let's look back.
First: find the height.
Then: write the rule.
Then: put the numbers in.
Then: work it out.
Last: check.