Question

identify the triangle that contains an acute angle for which the sine and cosine ratios are equal.

Explanation:

Step1: Recall the trigonometric identity

We know that \(\sin\theta=\cos(90^{\circ}-\theta)\). If \(\sin\theta = \cos\theta\), then \(\theta=90^{\circ}-\theta\). Solving the equation \(\theta + \theta=90^{\circ}\), we get \(2\theta = 90^{\circ}\), so \(\theta = 45^{\circ}\).

Step2: Analyze each triangle

• For the first triangle: The acute angles are \(40^{\circ}\) and \(50^{\circ}\). \(\sin40^{\circ}

eq\cos40^{\circ}\), \(\sin50^{\circ}
eq\cos50^{\circ}\).

• For the second triangle: It is a right - isosceles triangle (\(AC = BC\) as marked by the tick marks). The acute angles are \(45^{\circ}\) and \(45^{\circ}\). Let \(\theta = 45^{\circ}\), \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\) and \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\).
• For the third triangle: The acute angles are \(22^{\circ}\) and \(68^{\circ}\). \(\sin22^{\circ}

eq\cos22^{\circ}\), \(\sin68^{\circ}
eq\cos68^{\circ}\).

• For the fourth triangle: The acute angles are \(30^{\circ}\) and \(60^{\circ}\). \(\sin30^{\circ}=\frac{1}{2}\), \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\); \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), \(\cos60^{\circ}=\frac{1}{2}\).

Answer:

The second triangle (the right - isosceles triangle with \(45^{\circ}\) acute angles)