High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize High School Grade 11 Math Textbook > Trigonometry

Special Angles & Special Triangles (Degrees)

Reciprocal Trig RatiosPrevious Section
Trig Ratios for Angles Between 0° and 360°Next Section
Popular Courses
Find My Course
0:00 / 0:00

Special Angles & Special Triangles (Degrees)

There are 2 special triangles that can be used to find exact values (no decimals) of trig ratios.

30-60-90 Triangle

The30°- 60°- 90°30\degree \text{- }60\degree\text{- }90\degree triangle is half of an equilateral triangle with side length 2:

Wize Tip
Memorize these side lengths: 1,3,21, \sqrt{3}, 2
When written in increasing order, just like that, they match up with the angles across from them: 30°- 60°- 90°30\degree \text{- }60\degree\text{- }90\degree
Example
What is the exact value of sin⁡60°\sin{60\degree}?
sin⁡60°=OppHyp=32\sin{60\degree} = \dfrac{Opp}{Hyp} =\dfrac{\sqrt{3}}{2}


PAGE BREAK

45-45-90 Triangle

The45°- 45°- 90°45\degree \text{- }45\degree\text{- }90\degree triangle looks like a square with side length 1 that was cut along its diagonal:

Wize Tip
Any right triangle with two equal sides always has angles of 45°45\degree.
Example
What is the exact value of cos⁡45°\cos{45\degree}?
cos⁡45°=AdjHyp=12\cos{45\degree} = \dfrac{Adj}{Hyp} =\dfrac{1}{\sqrt{2}}
We like not to have roots in the denominator, so we can multiply top and bottom by 2\sqrt{2}:
cos⁡45°=12×22=22\cos{45\degree} =\dfrac{1}{\sqrt{2}} \colorFour{\times \small\dfrac{\sqrt{2}}{\sqrt{2}}} = \boxed{\dfrac{\sqrt{2}}{2}}

0:00 / 0:00

Example: Special Angles & Special Triangles (Degrees)

a) Evaluate the following expression exactly:

cos⁡(30°)+tan⁡2(60°)\cos(30\degree)+ \tan^2(60\degree)

Watch Out!
tan⁡2(60°)\tan^2(60\degree) is special notation that means "square this whole expression":
tan⁡2(60°)=[tan⁡(60°)]2\tan^2(60\degree)=\big[\tan(60\degree) \big]^2

The angles are 30°30\degree and 60°60\degree, so we draw the 30°- 60°- 90°30\degree \text{- }60\degree\text{- }90\degree triangle:

Let's find each trig ratio separately:

cos⁡(30°)=AdjHyp=32\hspace{2.4em} \cos(30\degree) = \dfrac{Adj}{Hyp} = \colorTwo{\dfrac{\sqrt{3}}{2}}


tan⁡(60°)=OppAdj=31=3  ⟹  tan⁡2(60°)=[tan⁡(60°)]2=[3]2=3\begin{aligned} &\tan(60\degree)=\dfrac{Opp}{Adj}=\dfrac{\sqrt{3}}{1} = \sqrt3\\[1em] \implies &\tan^2(60\degree) = \big[\tan(60\degree) \big]^2 = [\sqrt3]^2 = \colorTwo{3} \end{aligned}

Now that we have exact expressions, we can evaluate the whole expression:

cos⁡(30°)+tan⁡2(60°)=32+3=32+3×22=32+62=3+62\begin{aligned} \cos(30\degree)+ \tan^2(60\degree) &= \colorTwo{\dfrac{\sqrt3}{2}} + \colorTwo{3} \\[1em] &= \dfrac{\sqrt3}{2} + 3\times\scriptsize{\frac{2}{2}} \\[1em] &= \dfrac{\sqrt3}{2} + \dfrac{6}{2} \\[1em] &= \boxed{\dfrac{\sqrt3+6}{2}} \end{aligned}

PAGE BREAK
b) Determine the acute angle θ\theta given the equation:

2sin⁡θ=1\sqrt{2}\sin\theta =1

We can solve for sin⁡θ\sin\theta by dividing both sides by 2\sqrt2:

2sin⁡θ2=12  ⟹  sin⁡θ=12\begin{aligned} \dfrac{\cancel{\sqrt{2}}\sin\theta}{\cancel{\sqrt2}} &= \dfrac{1}{\sqrt2}\\[1em] \implies \sin\theta &= \dfrac{1}{\sqrt2} \end{aligned}

What special triangle has side-lengths of 11 and 2\sqrt{2}? ⟶45°- 45°- 90°\longrightarrow 45\degree \text{- }45\degree\text{- }90\degree

Now we can see that taking sin⁡45°\sin{45\degree} (from either of the angles in the triangle) gives 12\dfrac{1}{\sqrt2}.

Therefore, the angle θ=45°\boxed{\theta=45\degree}



Practice: Special Angles & Special Triangles

Evaluate the following expression, taking θ=60°\theta=60\degree:
cos⁡θ+sin⁡θ+tan⁡θ\cos\theta+\sin\theta+\tan\theta
Leave your answer in exact form.

Practice: Special Angles & Special Triangles

Determine the special angle θ\theta (in degrees) for each equation:

a) 6cos⁡θ=336\cos\theta=3\sqrt{3}

b) tan⁡θ=1\tan\theta=1