Triangle: Area (Using Sine)

Sides: 5, 6
Angle formed by these two sides: 45º

Then (area) = [1/2]⋅5⋅6⋅(sin 45º).

To find sin 45º,
draw a 45-45-90 triangle
whose sides are 1, 1, √2.

[1/2]⋅5⋅6 = 5⋅3

Find sin 45º.

Sine is SOH:
Sine,
Opposite side (1),
Hypotenuse (√2).

So sin 45º = 1/√2.

So [1/2]⋅5⋅6⋅(sin 45º)
= 5⋅3⋅[1/√2].

5⋅3 = 15

To rationalize the denominator √2,
multiply [√2/√2].

15√2 = 15√2

√2⋅√2 = 2

So 15√2/2 is the answer.

Sides: 4, 7
Angle formed by these two sides: 120º

Then (area) = [1/2]⋅4⋅7⋅(sin 120º).

To find sin 120º,
draw a terminal side
on a coordinate plane.

The reference angle of 120º is,
180 - 120, 60º.

Draw a right triangle.

The reference angle is 60º.
So the sides of the right triangle are
-1, √3, 2.

30-60-90 triangle

[1/2]⋅4⋅7 = 2⋅7

Find sin 120º.

Sine is SOH:
Sine,
Opposite side (√3),
Hypotenuse (2).

So sin 120º = √3/2.

Sine: Value

So [1/2]⋅4⋅7⋅(sin 120º)
= 2⋅7⋅[√3/2].

Cancel the common factor 2.

So 7√3 is the answer.