Question

For the square ABCD, it is known that | AB| = | AC| and |DA| = |DC| and corner ABC =

corner CDA. One of the corners of this square is 150 degrees. Find the sizes of the three remaining angles of the square ABCD.

Answer 1

• ∠BAD = 150°
• ∠ABC = 48°
• ∠BCD = 114°
• ∠ADC = 48°

Explanation:

Given quadrilateral ABCD with AB≅AC, DA≅DC, ∠ABC≅∠ADC, and one of the corner angles equal to 150°, you want the measures of all corner angles.

Setup

The given congruent angles and sides are marked in the attached diagram. If we let ∠ACB = x and ∠ACD = y, we can write some equations involving the sums of the various angles.

First of all, we note that both triangles are isosceles. That means ...

• ∠ABC≅∠ACB = x
• ∠DAC≅∠DCA = y

The sum of angles in ∆ACD must be 180°, so we have ...

x + 2y = 180°

We know that the isosceles triangle base angles cannot exceed 90°, so the angles at corners B and D cannot be 150°. That leaves two possibilities:

(a) Corner angle C is 150° ⇒ x + y = 150°

(b) Corner angle A is 150° ⇒ (180°-2x) +y = 150°

Solution

(a) Corner C is 150°

This makes the system of equations for x and y be ...

• x +2y = 180°
• x +y = 150°

Subtracting the second from the first gives y = 30°, and substituting that value for y gives x = 120°. We already know x is the base angle of an isosceles triangle, so cannot have that value. This possibility is eliminated.

(b) Corner D is 150°

This makes the system of equations for x and y be ...

• x +2y = 180°
• (180° -2x) +y = 150°

The second of these equations can be rearranged to ...

2x -y = 30°

Adding twice this to the first equation, we have ...

2(2x -y) +(x +2y) = 2(30°) +(180°)

5x = 240°

x = 48°

Substituting for x in the second equation gives ...

2(48°) -y = 30°

96° -30° = y = 66°

As we noted in the solution part (a), corner angle C is the sum of x and y:

∠BCD = x +y = 48° +66° = 114°

Corner angles

Then the corner angles are ...

• ∠BAD = 150°
• ∠ABC = 48°
• ∠BCD = 114°
• ∠ADC = 48°

The attached figure is drawn to scale.