Step by step
The question
Question: find the area of the triangle with sides 9 and 7, and the angle 50° 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 7 as its longest side, and h across from the 50° angle.
SOH: sin is the opposite over the hypotenuse.
sin 50° = h/7.
To remove the 7 at the bottom, we multiply by 7.
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.
7/7 is 1.
So just h is left.
So h = 7 × sin 50°.
2 Write the rule
The area of a triangle is a half times the base times the height.
The base is 9, the height is 7 × sin 50°.
So the rule is ½ × a × c × sin B: 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 9 and 7, and sin 50°.
4 Work it out
9 times 7 is 63.
Half of 63 is 31.5.
The calculator gives sin 50°: about 0.7660.
Put the number in.
That is about 24.13.
To the nearest tenth, the area is about 24.1 square units.
5 Check
Let's check another way: use the side 7 as the base.
Its height is 9 times sin 50°: about 6.89.
A half times 7 times that is about 24.1.
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.