Question

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C only. You will need to use both Soh-Cah-Toa and the Pythagorean Theorem to solve for side AB.

Answer 1

`AB = √(a^2 + b^2-2abCos\ C)`

Explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

`Sin\ C = (h)/(a)`

Make h the subject:

`h = aSin\ C`

Also, in BOC (Using Pythagoras)

`a^2 = h^2 + x^2`

Make
`x^2` the subject

`x^2 = a^2 - h^2`

Substitute
`aSin\ C` for h

`x^2 = a^2 - h^2` becomes

`x^2 = a^2 - (aSin\ C)^2`

`x^2 = a^2 - a^2Sin^2\ C`

Factorize

`x^2 = a^2 (1 - Sin^2\ C)`

In trigonometry:

`Cos^2C = 1-Sin^2C`

So, we have that:

`x^2 = a^2 Cos^2\ C`

Take square roots of both sides

`x= aCos\ C`

In triangle BOA, applying Pythagoras theorem, we have that:

`AB^2 = h^2 + (b-x)^2`

Open bracket

`AB^2 = h^2 + b^2-2bx+x^2`

Substitute
`x= aCos\ C` and
`h = aSin\ C` in
`AB^2 = h^2 + b^2-2bx+x^2`

`AB^2 = h^2 + b^2-2bx+x^2`

`AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2`

Open Bracket

`AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C`

Reorder

`AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C`

Factorize:

`AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C`

In trigonometry:

`Sin^2C + Cos^2 = 1`

So, we have that:

`AB^2 = a^2 * 1 + b^2-2abCos\ C`

`AB^2 = a^2 + b^2-2abCos\ C`

Take square roots of both sides

`AB = √(a^2 + b^2-2abCos\ C)`