Solving expressions using 30-60-90 special right triangles

Everything You Need in One Place

Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.

Learn and Practice With Ease

Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.

Instant and Unlimited Help

Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!

  1. What are special triangles?
  1. If θ=30 \theta = 30^\circ , find the exact value of the following expressions.
    1. sin2θ \sin 2 \theta
    2. cos22θ \cos^2 2 \theta
    3. tan2θ2{{\tan^2 \theta} \over 2}
  2. If θ=60 \theta = 60^\circ , find the exact value of the following expressions.
    1. 4sin3θ2 4 \sin^3 {\theta \over 2}
    2. 23cos2θ2{2 \over {{3 \cos^2}{ \theta \over 2}}}
    3. 3tan3θ23 \tan^3 {\theta \over 2}
Topic Notes

What is a 30-60-90 triangle?

One of the two special right triangles you'll be facing in trigonometry is the 30-60-90 triangle. The other one is the 45 45 90 triangle. These triangles are special triangles because the ratio of their sides are known to us so we can make use of this information to help us in right triangle trigonometry problems.

In the case of the 30-60-90 triangle, their side's ratios are 1 : 2 : 3\sqrt3. We'll prove that this is true first so that you can more easily remember the triangle's properties.

A 30-60-90 triangle is actually half of an equilateral triangle.

Triangle ABC shown here is an equilateral triangle. Since it's equilateral, each of its 3 angles are 60 degrees respectively. Its sides are also equal.

If we draw a line AD down the middle to bisect angle A into two 30 degree angles, you can now see that the two new triangles inside our original triangle are 30-60-90 triangles.

We know that BD is equalled to DC, which is also equalled to half of AB since AB was originally equalled to BC before it was split in half. So therefore, the ratio of BD : AB is 1 : 2.

Since we have a hypotenuse due to the 90 degree angle that exists inside our triangle, we can use the pythagorean theorem to figure out the last side: AD.

12+AD2=221^2 + AD ^2 = 2^2

AD2=41AD ^2 = 4 - 1

AD2=3AD^2 = 3

AD=3AD = \sqrt 3

And there you have it! You now know how all the sides of the 30-60-90 triangles came about. Let's move on to solving right triangles with our knowledge on the sides' ratios.

Solve for 30 60 90 triangle

Here we have a 30-60-90 special right triangle, with the three interior angles of 30, 60, 90 degrees. We know that the length of each side in this triangle is in a fixed ratio. We can now use the ratio to solve the following problem.

Question 1:

If   Θ = 30° , find the exact value of the following expression:

sin 2Θ


Since Θ = 30°, sin 2Θ = sin 60°.

We know that sin Θ = oppositehypotenuseopposite \over hypotenuse

Since we are looking for the value of sin 60°, we are going to look at the 60° angle in the special triangle and find out the opposite side and hypotenuse of the angle. And we found that the opposite side = 3 and the hypotenuse = 2. Then, we put the numbers into sin 60° = oppositehypotenuseopposite \over hypotenuse. This gives us the final answer of 32{\sqrt3}\over2

Question 2:

If  Θ=60° , find the exact value of the following expression:

4sin34sin^3θ2{{\theta} \over 2}


First, we put Θ = 60° into 4sin34sin^3θ2{{\theta} \over 2} . We get

4sin3sin^3(602)({60 \over 2}) = 4sin34sin^3(30°)

From the special triangle, we know that sin 30° = 121\over2

Then, we can now solve the cubic root:


The final answer is 121\over2

A good online resource to help you check your work when it comes to 30-60-90 triangles (or even 45 45 90 triangles!) can be found here. You can also play along with this online scalable 30-60-90 triangle to see how the sides change while staying in the 1 : 2 : 3\sqrt3 ratio.

sin30=12 \sin 30^\circ = { 1 \over 2}
cos30=32 \cos 30^\circ = {{\sqrt 3} \over 2}
tan30=13 \tan 30^\circ = {1 \over \sqrt 3}
sin60=32 \sin 60^\circ= {{\sqrt3} \over 2}
cos60=12 \cos 60^\circ= {1 \over 2}
tan60=31 \tan 60^\circ = {{\sqrt 3} \over 1}

angles and side ratios in a 30-60-90 special right triangle