Strange angles

Geometry Level 2

A B AB is parallel to C D CD and A B = 2 C D AB =2CD .

If m A B C = 4 5 o m\angle ABC = 45^o and m C B D = 1 5 o , m\angle CBD = 15^o, find m B A D m\angle BAD in degrees.


The answer is 75.

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5 solutions

B C D = A B C = 4 5 \angle BCD=\angle ABC=45^\circ

Apply sine rule on B C D \triangle BCD :

B D sin 45 = 1 sin 15 \dfrac{BD}{\sin 45}=\dfrac{1}{\sin 15} \implies B D = sin 45 sin 15 BD=\dfrac{\sin 45}{\sin 15}

Apply cosine rule on A B D \triangle ABD :

A D 2 = 2 2 + ( sin 45 sin 15 ) 2 2 ( 2 ) ( sin 45 sin 15 ) ( cos 60 ) AD^2=2^2+\left(\dfrac{\sin 45}{\sin 15}\right)^2-2(2)\left(\dfrac{\sin 45}{\sin 15}\right)(\cos 60)

A D = 6 AD=\sqrt{6}

Apply sine rule on A B D \triangle ABD :

sin B A D sin 45 sin 15 = sin 60 6 \dfrac{\sin \angle BAD}{\dfrac{\sin 45}{\sin 15}}=\dfrac{\sin 60}{\sqrt{6}}

B A D = 7 5 \angle BAD = \color{#D61F06}\boxed{75^\circ}

Steve Shaff
Nov 11, 2017

Let F be the foot of the perpendicular from A to BD and let Q be the foot of the perpendicular from B to CD extended through D. Since triangle ABF is 30-60-90, it follows that BF = x and AF = x sqrt 3. Let DQ = y. Since parallel lines AB and CD are cut by transversals BC and BD, alternate interior angles are congruent so triangle BDQ and triangle BCQ are 30-60-90 and 45-45-90 respectively. Let DQ = y. It now follows that BD = 2y, BQ = y sqrt 3, and x + y = CD + DQ = CQ = BQ = y sqrt 3. so that y = x/(sqrt 3 - 1) = x (sqrt 3 + 1)/2 and BD = 2y = x(sqrt 3 +1). Next note that DF = BD - BF = x(sqrt 3 + 1) - x = x sqrt 3 = AF. Therefore triangle AFD is 45-45-90 with angle DAF = 45. Finally, we have angle DAB = angle DAF + angle FAB = 45 + 30 = 75.

Henry Carpenter
Nov 14, 2017

O is the centre. Triangles OCA and OBD must be similar. Therefore angle CAO is 15. Then because angle ABC is 45, the whole thing must be a chunk of a square, so angle CAB must be 90. Therefore the remainder is 90 minus 15= 75

I doubt if one can be 'sure' that its a chunk of a square, just because of 45 degree angle

Rupesh Gesota - 3 years, 6 months ago

CAB is NOT 90 degrees. The 2 triangles you mention are NOT similar at all. Good thought though.

Daniel Ettedgui - 2 years, 5 months ago

ΔOCA and ΔOBD are not similar! What do you mean “they must be similar”? But ΔOAB and ΔOCD are always guranteed similar

Ավո Ակիլյան - 2 years, 3 months ago

Where did the O in CAO and OCA and OBD come from?

Christine Tan - 11 months, 2 weeks ago
Tudor Cotitosu
Apr 13, 2020

link text We have some relations between an angle form the center of a circle and an angle form a point on the circle for a given arc! The angle form the center of the circle is always duble than the angle form a point on the circle!

Daniel Ettedgui
Dec 26, 2018

Since triangles AOB and OCD are similar, then their altitudes are also in a ratio of 2 to 1. If smaller triangle height is h, then larger triangle height is 2h. Drop a perpendicular from B to CD extended. This creates a 30/60/90 triangle with height 3h and base h(sqroot3). CD extended is also 3h since the diagonal creates a 45/45/90 isosceles triangle.
Now Tan theta is h/(CD-h). Another way to write CD is 3h-h(sqroot3). Look at that, tan theta = 2+sqroot3. Theta is therefore 75 degrees!

Are you sure this works

Daksh Chauhan - 2 years, 5 months ago

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