Angle of a trapezium

Geometry Level 5

A B C D ABCD is a trapezium with A D ∥ B C AD \parallel BC and A B = A D AB = AD . 0 ∘ < θ < 18 0 ∘ 0^\circ < \theta < 180^\circ is an angle such that ∠ B A C = θ \angle BAC = \theta , ∠ C A D = 3 θ \angle CAD = 3 \theta and ∠ A C D = 5 θ \angle ACD = 5 \theta . What is the measure (in degrees) of θ \theta ?

Details and assumptions

A trapezium has a pair of parallel sides.


The answer is 10.

Calvin Lin by Calvin Lin

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2 solutions

Aradhya Kasera
Aug 6, 2013

Note all angle measures are in degrees

Let θ = a \theta=a for convenience.

Now for some housekeeping: Let A B = A D = s AB=AD=s . Since A B C D ABCD is a trapezoid, ∠ A = a + 3 a = 4 a \angle A=a+3a=4a so ∠ B = 180 − 4 a \angle B=180-4a . Since ∠ B A C = a \angle BAC =a , we have ∠ B C A = 180 − ( a + 180 − 4 a ) = 3 a \angle BCA=180-(a+180-4a)=3a , then ∠ A D C = 180 − 3 a − 5 a = 180 − 8 a \angle ADC=180-3a-5a=180-8a . Since 180 − 8 a 180-8a must be positive, we have a < 22.5 a<22.5 . Now we have a new restraint, 0 < a < 22.5 0<a<22.5 .

Applying the Law of Sines to △ A D C \triangle ADC we see sin ⁡ ( 180 − 8 a ) A C = s i n 5 a s \frac{\sin(180-8a)}{AC}=\frac{sin5a}{s} .

Again applying the Law of Sines to △ A B C \triangle ABC results in sin ⁡ ( 180 − 4 a ) A C = s i n 3 a s \frac{\sin(180-4a)}{AC}=\frac{sin3a}{s} .

Equating the ratio A C s = sin ⁡ ( 180 − 4 a ) sin ⁡ 3 a = sin ⁡ ( 180 − 8 a ) sin ⁡ 5 a \frac{AC}{s}=\frac{\sin(180-4a)}{\sin3a}=\frac{\sin(180-8a)}{\sin5a} .

Since sin ⁡ ( 180 − θ ) = sin ⁡ θ \sin(180-\theta)=\sin\theta , we have sin ⁡ ( 4 a ) sin ⁡ 3 a = sin ⁡ ( 8 a ) sin ⁡ 5 a \frac{\sin(4a)}{\sin3a}=\frac{\sin(8a)}{\sin5a} .

Note that sin ⁡ 8 a = 2 sin ⁡ 4 a cos ⁡ 4 a \sin8a=2\sin4a\cos4a , so we have sin ⁡ 5 a sin ⁡ 3 a = 2 cos ⁡ 4 a \frac{\sin5a}{\sin3a}=2\cos4a or sin ⁡ 5 a = 2 cos ⁡ 4 a sin ⁡ 3 a \sin5a=2\cos4a\sin3a .

Using the identity sin ⁡ θ − sin ⁡ γ 2 = sin ⁡ θ − γ 2 cos ⁡ θ + γ 2 \frac{\sin\theta-\sin\gamma}{2} = \sin\frac{\theta-\gamma}{2}\cos\frac{\theta+\gamma}{2} we obtain sin ⁡ 5 a = sin ⁡ 7 a − sin ⁡ a \sin5a=\sin7a-\sin a or sin ⁡ a = sin ⁡ 7 a − sin ⁡ 5 a \sin a=\sin7a-\sin5a

Applying the same identity to the RHS of this new equation, results in 2 sin ⁡ a cos ⁡ 6 a = sin ⁡ a 2\sin a\cos6a=\sin a .

Clearly sin ⁡ a ≠ 0 \sin a \neq 0 , so we have cos ⁡ 6 a = 1 2 \cos6a=\frac{1}{2} , 6 a = 60 ± 360 n 6a=60\pm360n or 6 a = − 60 ± 360 n 6a=-60\pm360n .

But by the restraint 0 < a < 22.5 0<a<22.5 , a = 10 a=\boxed{10} is the only possible answer.

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

What happens to the diagram in the other solutions where a > 22.5 a > 22.5 ? Can you tie an explanation without drawing an accurate diagram?

Oops, a typo. It should be sin ⁡ a ≠ 0 \sin a\neq0

Aradhya Kasera - 7 years, 10 months ago

I believe that the resulting shape would not be a polygon, let alone a trapezoid. It would have 1 or more unconnected sides.

Aradhya Kasera - 7 years, 10 months ago
Daniel Wang
Aug 6, 2013

Let A B = A D = x AB=AD=x and A C = y AC=y . By Law of Sines on Δ A B C \Delta ABC ,

sin ⁡ 3 θ x = sin ⁡ ( 180 − 4 θ ) y ⟹ sin ⁡ 3 θ sin ⁡ ( 180 − 4 θ ) = x y \frac{\sin3\theta}{x}=\frac{\sin(180-4\theta)}{y} \implies \frac{\sin3\theta}{\sin(180-4\theta)}=\frac{x}{y}

By Law of Sines on Δ A C D \Delta ACD ,

sin ⁡ 5 θ x = sin ⁡ ( 180 − 8 θ ) y ⟹ sin ⁡ 5 θ sin ⁡ ( 180 − 8 θ ) = x y \frac{\sin5\theta}{x}=\frac{\sin(180-8\theta)}{y} \implies \frac{\sin5\theta}{\sin(180-8\theta)}=\frac{x}{y}

Therefore, sin ⁡ 3 θ sin ⁡ ( 180 − 4 θ ) = sin ⁡ 5 θ sin ⁡ ( 180 − 8 θ ) \frac{\sin3\theta}{\sin(180-4\theta)}=\frac{\sin5\theta}{\sin(180-8\theta)} , and we simplify.

sin ⁡ 3 θ sin ⁡ ( 4 θ ) = sin ⁡ 5 θ sin ⁡ ( 8 θ ) \frac{\sin3\theta}{\sin(4\theta)}=\frac{\sin5\theta}{\sin(8\theta)}

sin ⁡ 3 θ sin ⁡ 8 θ = sin ⁡ 4 θ sin ⁡ 5 θ \sin3\theta\sin8\theta=\sin4\theta\sin5\theta

sin ⁡ 3 θ ( 2 cos ⁡ 4 θ sin ⁡ 4 θ ) = sin ⁡ 4 θ sin ⁡ 5 θ \sin3\theta(2\cos4\theta\sin4\theta)=\sin4\theta\sin5\theta

sin ⁡ 3 θ ( 2 cos ⁡ 4 θ ) = sin ⁡ 5 θ \sin3\theta(2\cos4\theta)=\sin5\theta

Luckily, I had sloppy handwriting, and mistaken θ \theta as 10 10 . Then I get:

sin ⁡ 30 ( 2 cos ⁡ 40 ) = sin ⁡ 50 \sin30(2\cos40)=\sin50

sin ⁡ 30 ( 2 sin ⁡ 50 ) = sin ⁡ 50 \sin30(2\sin50)=\sin50

sin ⁡ 30 = 1 2 \sin30=\frac{1}{2}

And then I thought, "Agh, I forgot the thhheta!" and changed the equation back to

sin ⁡ 3 θ = 1 2 \sin3\theta=\frac{1}{2}

and so, the answer is 10 \boxed{10}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This is a common mistake made in trigonometric solutions, which forget that these functions are not one-to-one:

  1. There are numerous solution to sin ⁡ 3 θ = 1 2 \sin 3 \theta = \frac{1}{2} , and you have not explained why the answer must only be 10.

You made my day with your mistake. Gotta love math humor :)

Ivan Sekovanić - 7 years, 10 months ago

i can explain that . sin 3theta =1/2 for 3theta=30,150 ,if the value of 3theta =150 then angle bcd would become 400 and we know that the sum of all the angles of a quadrilateral is 360

SHASHANK GOEL - 7 years, 10 months ago

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