In a right triangle three ratios are defined with respect to an acute angle \(\alpha\):

The hypotenuse is the side opposite the right angle (always the longest). The opposite leg is the side opposite angle \(\alpha\); the adjacent leg is the side next to \(\alpha\) that is not the hypotenuse.

What happens in quadrilaterals?

• In a rectangle all angles are \(90^\circ\). One diagonal divides the rectangle into two congruent right triangles. In such a triangle the diagonal is always the hypotenuse, and the two sides of the rectangle are the legs.
• Two diagonals of a rectangle intersect and create four triangles — but these are not right triangles (the diagonals of a rectangle are not perpendicular).
• In a rhombus the diagonals are perpendicular to each other and bisect each other, so they create four right triangles.
• In a parallelogram, to obtain a right triangle you drop an altitude from a vertex to a side (or its extension).

Example 1: Identifying Sine in a Triangle Formed by a Diagonal in a Rectangle

Problem: In rectangle \(ABCD\) diagonal \(BD\) is drawn. In triangle \(ABD\) the right angle is at \(A\), and \(AB = 6\) cm, \(AD = 8\) cm. Find \(\sin(\angle ADB)\).

Solution:

1. In triangle \(ABD\) the right angle is at \(A\), so diagonal \(BD\) is the hypotenuse.
2. Calculate the hypotenuse using the Pythagorean Theorem: \(BD = \sqrt{AB^2 + AD^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\).
3. With respect to angle \(\angle ADB\), the opposite leg is \(AB = 6\) (the side opposite angle \(D\)).
4. Therefore \(\sin(\angle ADB) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{AB}{BD} = \dfrac{6}{10} = 0.6\).

Answer: \(\sin(\angle ADB) = \dfrac{6}{10} = 0.6\).

Example 2: Cosine and Tangent for the Same Angle

Problem: In rectangle \(ABCD\) diagonal \(BD\) is drawn, with the right angle of triangle \(BCD\) at \(C\). Given \(BC = 5\) cm and \(CD = 12\) cm, find \(\cos(\angle BDC)\) and \(\tan(\angle BDC)\).

Solution:

1. The right angle is at \(C\), so \(BD\) is the hypotenuse: \(BD = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\).
2. With respect to angle \(\angle BDC\) (at vertex \(D\)): the adjacent leg is \(CD = 12\) and the opposite leg is \(BC = 5\).
3. \(\cos(\angle BDC) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{CD}{BD} = \dfrac{12}{13}\).
4. \(\tan(\angle BDC) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{BC}{CD} = \dfrac{5}{12}\).

Answer: \(\cos(\angle BDC) = \dfrac{12}{13}\), \(\tan(\angle BDC) = \dfrac{5}{12}\).

Example 3: Identifying the Hypotenuse in a Triangle from a Parallelogram with an Altitude

Problem: In parallelogram \(ABCD\) altitude \(DH\) is dropped from vertex \(D\) to side \(AB\), forming a right angle at \(H\). In triangle \(AHD\), given \(AD = 10\) cm and \(\angle DAH = 39^\circ\), express \(DH\) using the appropriate trigonometric ratio.

Solution:

1. In triangle \(AHD\) the right angle is at \(H\), so \(AD\) (the side of the parallelogram) is the hypotenuse.
2. With respect to angle \(\angle DAH = 39^\circ\): the altitude \(DH\) is the leg opposite the angle.
3. Therefore \(\sin(39^\circ) = \dfrac{DH}{AD}\), giving \(DH = AD\cdot\sin(39^\circ) = 10\cdot\sin(39^\circ)\).
4. Calculating: \(\sin(39^\circ)\approx 0.629\), so \(DH \approx 6.29\) cm.

Answer: \(DH = 10\sin(39^\circ) \approx 6.29\) cm.

Example 4: Perpendicular Diagonals in a Rhombus

Problem: In rhombus \(ABCD\) the diagonals intersect at point \(O\). Given half-diagonal \(AO = 8\) cm and half-diagonal \(BO = 6\) cm, find \(\sin(\angle BAO)\) in triangle \(ABO\).

Solution:

1. In a rhombus the diagonals are perpendicular, so the angle at \(O\) is \(90^\circ\) and triangle \(ABO\) is a right triangle.
2. Side \(AB\) (a side of the rhombus) is the hypotenuse: \(AB = \sqrt{AO^2 + BO^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\).
3. With respect to angle \(\angle BAO\) (at vertex \(A\)): the opposite leg is \(BO = 6\).
4. Therefore \(\sin(\angle BAO) = \dfrac{BO}{AB} = \dfrac{6}{10} = 0.6\).

Answer: \(\sin(\angle BAO) = \dfrac{6}{10} = 0.6\).

✗ Common mistake: Assuming that the diagonals of a rectangle are perpendicular and claiming that four right triangles are formed.

✓ The correct way: In a rectangle the diagonals are equal but not perpendicular, so the four triangles they create are not right triangles. Perpendicular diagonals occur in a rhombus (and in a square).

✗ Common mistake: Mixing up the opposite leg and the adjacent leg relative to the angle, and using sine instead of cosine.

✓ The correct way: Always mark the specific angle: \(\sin\) measures the side opposite the angle divided by the hypotenuse; \(\cos\) measures the side adjacent to the angle divided by the hypotenuse.

✗ Common mistake: Treating the diagonal as a leg in the sine formula for a triangle formed in a rectangle.

✓ The correct way: The diagonal lies opposite the right angle, so it is always the hypotenuse. The legs are the sides of the quadrilateral.

The ratios: \(\sin\alpha = \dfrac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\alpha = \dfrac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\alpha = \dfrac{\text{opposite}}{\text{adjacent}}\).

• The diagonal in a rectangle is the hypotenuse; the sides of the rectangle are the legs.
• Rectangle: one diagonal creates 2 right triangles.
• Rhombus: two perpendicular diagonals create 4 right triangles.
• In a parallelogram, drop an altitude to obtain a right triangle.
• Always identify the hypotenuse (opposite the right angle) before applying any ratio.