Step by step
The question
Question: find the area of the triangle with sides 7 cm and 10 cm, and the angle 45° 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 10 as its longest side, and h across from the 45° angle.
SOH: sin is the opposite over the hypotenuse.
sin 45° = h/10.
To remove the 10 at the bottom, we multiply by 10.
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.
10/10 is 1.
So just h is left.
So h = 10 × sin 45°.
2 Write the rule
The area of a triangle is a half times the base times the height.
The base is 7, the height is 10 × sin 45°.
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 7 and 10, and sin 45°.
4 Work it out
7 times 10 is 70.
Half of 70 is 35.
The calculator gives sin 45°: about 0.7071.
Put the number in.
That is about 24.749.
To the nearest tenth, the area is about 24.7 cm².
5 Check
Let's check another way: use the side 10 as the base.
Its height is 7 times sin 45°: about 4.95.
A half times 10 times that is about 24.7.
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.