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  • 7 Hard Trig Problems—Can You Solve These Challenging Geometry Problems?

7 Hard Trig Problems—Can You Solve These Challenging Geometry Problems?

Updated on December 22, 2016
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TR Smith is a teacher and creator who uses mathematics in her line of work every day.

Trigonometry is an area of geometry that deals with lengths and angles of triangles. And since many two-dimensional figures and three-dimensional sold shapes can be broken down into triangles, trigonometry is also used to solve many advanced problems involving more complicated shapes.

Trigonometric relations and functions are based on the relations between the sides of right triangles. The sine of a non-right angle is the ratio of the opposite side to the hypotenuse. The cosine is the ratio of the adjacent side to the hypotenuse. And finally, the tangent is the ratio of the opposite side to the adjacent side. Many people use the mnemonic "SohCahToa" to remember these ratios. Try these seven hard geometry problems to test your skills at using trig to find angles and lengths in triangles.

(1) Right Triangles of Equal Perimeter

Problem: The two right triangles shown above have equal perimeters. The hypotenuse of the orange triangle is one leg of the green triangle stacked on top of it. If the smallest angle of the orange triangles is 20 degrees, what are the angles of the green right triangle?

Solution: We don't need to know the side lengths of the triangles to solve for the unknown angles, only their relative lengths are necessary. For the sake of simplicity, let the two legs of the orange triangle be cos(20) and sin(20), and let the hypotenuse equal 1. The green triangle then has one leg of length 1, another length of unknown length X, and a hypotenuse of sqrt(X^2 + 1).

Since the two triangles' perimeters are equal, we have the equation

sin(20) + cos(20) + 1 =
1 + X + sqrt(X^2 + 1)

sin(20) + cos(20) =
X + sqrt(X^2 + 1)

If we use the technique for solving algebraic equations with square root and linear terms, we get

[sin(20) + cos(20) - X]^2 = X^2 + 1

sin(20)^2 + 2sin(20)cos(20) + 2cos(20)^2 - 2sin(20)X - cos(20)X + X^2
= X^2 + 1

1 + 2sin(20)cos(20) - 2sin(20)X - 2cos(20)X + X^2
= X^2 + 1

2sin(20)cos(20) - 2sin(20)X - 2cos(20)X = 0

X = sin(20)cos(20)/[sin(20 + cos(20)] ≈ 0.250753

The angles of the green triangle are 90 degrees, arctan(X/1) ≈ 14.08 degrees, and arctan(1/X) ≈ 75.92 degrees.

(2) Find the Missing Angles

Problem: Consider a triangle with one angle equal to 45 degrees. An altitude is drawn to partition the original triangle into two smaller right triangles, colored yellow and purple in the diagram above. If the area of the purple triangle is 1.5 times the area of the yellow triangle, what are the angles of the original whole triangle?

Solution: An altitude line splits the triangle into two right triangles. The yellow triangle is a 45-45-90-degree triangle, so its legs have equal length. Without any loss of generality, we can set that length equal to 1. If the purple triangle has 1.5 times the area, then its leg lengths are 1 and 1.5.

Call the top unknown angle of the purple triangle x degrees and the bottom unknown angle (90-x) degrees. From trigonometry, we know that

tan(x) = 1.5/1 = 1.5

x = arctan(1.5) ≈ 56.31 degrees

Since 90 - 56.31 = 33.69 and 45 + 56.31 = 101.31, we now have all the angles of the original triangle: 45°, 33.69°, and 101.31°.

(3) Isosceles Integer Triangles

Problem: An isosceles triangle has a perimeter of 18 and integer side lengths. What is the smallest possible area and what is the smallest possible angle measure at one of its corners?

Solution: There are only a finite number of triangles that have integer side lengths, are isosceles, and have a perimeter of 18. They are:

  • {5, 8, 5}
  • {6, 6, 6}
  • {7, 4, 7}
  • {8, 2, 8}

Using the Pythagorean Theorem we can find the heights of the triangles when the base is the unequal side. The area of each triangle is one half of the base times height. These computations work out to:

  • {5, 8, 5}
    height = 3
    area = 12
  • {6, 6, 6}
    height = 3*sqrt(3)
    area = 9*sqrt(3)
  • {7, 4, 7}
    height = 3*sqrt(5)
    area = 6*sqrt(5)
  • {8, 2, 8}
    height = 3*sqrt(7)
    area = 3*sqrt(7)

Of these, the smallest area is the last.

The two candidates for having the smallest angle are the triangles with sides {5, 5, 8} and {8, 8, 2}, the triangles at the extreme ends of the list.

Drawing the altitudes of the {5, 5, 8} triangle, we can see that its acute angle θ is equal to arctan(3/4) ≈ 36.87 degrees.

Drawing the altitude of the {8, 8, 2} triangle, we can see that its smallest acute angle φ satisfies the relation tan(φ/2) = 1/sqrt(63), and therefore φ is equal to 2*arctan(1/sqrt(63)) ≈ 14.36 degrees. This means the {8, 8, 2} triangle has the smallest angle.

You can verify that the angles of the {6, 6, 6} and {7, 7, 4} triangles are all larger than 14.36 degrees. In fact, the {6, 6, 6} triangle is equilateral, so its angles are all 60 degrees.

(4) Bisected Angle

Problem: Consider a scalene right triangle such that if you draw a line that bisects the mid-sized angle, you end up with two smaller triangles, one of which has three times the area of the other. What are the angles of this triangle?

Solution: Though no absolute lengths are given in this problem, we can solve it by using relative lengths. Let's set the length of the shorter leg of the large right triangle equal to 1. If the leg of the small right triangle is x and the other sub-triangle has three times the area, then its other side has a length of 3x (as implied by the base-height formula for the area of a triangle, since the heights are equal).

If the line segment indicated by the red arrow is an angle bisector, then the angle θ satisfies the two trigonometric relationships

tan(θ) = x, and

tan(2θ) = 4x

Eliminating the variable x gives us

tan(2θ) = 4tan(θ)

Here we can use the double angle trig identity tan(2θ) = 2tan(θ)/[1 - tan(θ)^2], which will give us an equation only in the variable "tan(θ)." Plugging this identity into the equation gives us

2tan(θ)/[1 - tan(θ)^2] = 4tan(θ)

1/[1 - tan(θ)^2] = 2

1 = 2 - 2tan(θ)^2

1 = 2tan(θ)^2

1/sqrt(2) = tan(θ)

θ = arctan(1/sqrt(2)) ≈ 35.26 degrees

The angles of the larger right triangle are 90 degrees, 2θ ≈ 70.53 degrees, and (90-2θ) ≈ 19.47 degrees.

(5) Angle at the Apex of a Cone

Problem: A quarter circular sector or is removed from a circle and the remainder is folded into a cone by connecting the cut edges. When viewed from the side, what is the angle at the apex of the cone?

Solution: Let the radius of the original circle be R. When the cone is folded from the 3/4 circular sector, the circumference of the cone's base is (3/4)*2*pi*R, which implies that the cone's base radius is (3/4)*R. The side length of the cone is R. Using the Pythagorean Theorem, the height of the cone is

sqrt[ R^2 - (9/16)*R^2 ] = R*sqrt[7/16] = R*sqrt(7)/4

Looking at the cone from the side, the side length is the hypotenuse of a right triangle with the cone radius and cone height as legs. If we call the angle at the top θ, then θ satisfies the trigonometric relation

sin(θ/2) = (3R/4) / R = 3/4

Solving this for θ gives us the solution θ = 2*arcsin(3/4) ≈ 97.18 degrees.

(6) Maximum Area of a Triangle

Problem: Two sides of a triangle have lengths 5 and 7. What should the length of the third side be so that the area of the triangle is maximized?

Solution: At first glance, this optimization problem might seem like one that should be solved with calculus rather than elementary trigonometry. If we call the missing side X, then Heron's Formula tells us that the area of the triangle is a function A(x) given by the equation

A(x) = 0.25*sqrt[ 148*X^2 - X^4 - 576 ]

The optimal value of X is obtained by taking the derivative of A(X), setting it equal to zero, and solving for X. However, there is a simpler trigonometric solution to this problem. Recall the triangular area formula for a triangle with two sides of length P and Q and an angle of θ between them:

Area = 0.5*P*Q*sin(θ)

In our problem, P = 5 and Q = 7, the only variable quantity in this equation is the angle θ, which can range from 0 degrees to 180 degrees. The area is then (35/2)*sin(θ). This area function is maximized when sin(θ) is maximized, and the maximum value of sin(θ) occurs at θ = 90 degrees where sin(θ) = 1.

If θ = 90 degrees, then the triangle is a right triangle whose legs are 5 and 7 and whose hypotenuse is sqrt(74). Thus, the missing side length should be sqrt(74) ≈ 8.6023 to achieve the maximum area. The area of this triangle is 0.5*5*7 = 35/2 = 17.5.

(7) Right Triangles of Equal Area

Problem: Two right triangles share a common side and have equal areas. The hypotenuse of the light green triangle is one of the legs of dark green triangle stacked on top of it. If the smallest angle of light green triangle is 18 degrees, what are the angles of the dark green triangle?

Solution: Hearkening back to the first problem in this set, the absolute lengths of the sides are not necessary, only the relative lengths. Let's say the legs of the light green triangle are cos(18) and sin(18) and the hypotenuse is 1. Then one leg of the dark green triangle is 1, the other leg is an unknown value X, and the hypotenuse is sqrt(X^2 + 1).

If two right triangles have equal areas, then the products of their respective legs are equal. Thus we have the relation

sin(18)*cos(18) = X*1

X ≈ 0.2938926

The angles of the dark green triangle are then 90 degrees, arctan(X/1) ≈ 16.38 degrees, and arctan(1/X) ≈ 73.62 degrees. Algebraically, this problem was much easier to solve than the first.


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    • 1701TheOriginal profile image

      Leonard Kelley 2 years ago

      Always great to see challenging math problems, helps build analytical skills. Great hub.

    • Joe 10 months ago

      I'm pretty sure there's an easier way to do the first problem

    • calculus-geometry profile image

      TR Smith 10 months ago from Germany

      If you have found a shorter solution please do share!

    • Joe 10 months ago

      sin(20)cos(20) is not the same as sin(20)cos(20)/[sin(20 + cos(20)],

      so the X's are not equal in both methods.

    • calculus-geometry profile image

      TR Smith 10 months ago from Germany

      Hi Joe, thanks for your comment.

      When solving for unknown values it is common to use "X" as a placeholder. You can use "X" to solve different problems, and doing so does not imply that the value of "X" is the same in these different problems. In particular, the unknown value represented by "X" in the first problem is not the same as in the last problem. This is standard algebra problem solving notation. (In number theory, for example, it is common to use m, n, k, p, and q for arbitrary integer values.)

      You may have noticed that the first and last problems look very very similar, so perhaps it is confusing to you that the first problem has a longer solution than the last. That is what makes them (as a pair) challenging geometry problems. People often assume their solutions should be similar. But since one is about perimeter and the other about area, you need to apply different concepts, which makes their solutions different.

    • Joe 10 months ago

      Hello, thanks for replying.

      The I didn't mean the the the solution for the X's in the first and last problem should be the same value. Of course I understood the solutions of the two problems are different, but the two problems are asking for the same thing, just with different angles: one is an angle of 20 degrees, the other is 18 degrees.

      As you mentioned after the solution of the last problem that "algebraically, this problem was much easier to solve than the first", it was obvious was easier to solve. However, the easier solution is questionable, making us wonder why so. If you were to use the same angle in the first problem for the last problem, you would get a different solution than the first problem. And the issue is the difference of the two methods, in which you don't get the same X's which I specify that

      sin(20)cos(20) is not the same as sin(20)cos(20)/[sin(20 + cos(20)].

    • Joe 10 months ago

      One other thing, I know you were using the perimeter to solve for the first and area to solve last problem. But you were still solving for the length of the shorter leg, X, WHILE the longer leg is still 1, which by my calculations the X's using area and perimeter weren't equal.

      For the angle = 20 degrees, using perimeter method X ≈ 0.250753

      using area method X ≈ 0.321394

    • calculus-geometry profile image

      TR Smith 10 months ago from Germany

      Hi Joe, I think I see what is confusing you. It seems you are considering the first and last problems to be the same problem but solved with two different "methods." They are actually two different problems each with its own method.

      It's not that the first problem is being solved with "the method of equating perimeters" and the last problem is being solved with "the method of equating areas." It's that the first problem is asking you to find an angle with a certain condition (same perimeter), and the second problem is asking you to find an angle with an entirely different condition (same area). This makes the two problems distinct. That is why the method used to solve one does not work for the other.

      If the angle in the last problem were 20 degrees instead of 18, you would indeed get two different answers for the two different problems. That isn't a mistake as you imply. That is a natural consequence of the fact that each constraint (perimeter and area) creates a separate problem.

      Here's an example of what I'm trying to explain in the same vein as the right triangle problem pair, but easier to work out in your head:

      Suppose you have a 2-by-8 rectangle. Problem A is to find the dimensions of a square with the same perimeter as this rectangle. Problem B is to find the dimensions of a square with the same area as this rectangle. What answers do you get for Problems A and B and how do you arrive at these answers? Do you see why the answers are different, even though the statements of the problems are superficially very similar?

    • Mann 9 months ago

      Nice work

    • Doug1943 5 months ago

      All of these are absolutely wonderful problems. And I also appreciate the author's tact and patience in answering questions from those (of whom I am one) who may find the going a bit hard.

      That's exactly the attitude we must have in the classroom, to encourage, not discourage, our students from asking questions which could, in the hands of a mathematically-able but psychologically-disabled teacher, be the occasion for student humilation. So a double-award to the author.

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