Question

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

Explanation:

Step1: Recall the property of sine and cosine

We know that for an acute angle \(\theta\), \(\sin\theta=\cos(90^{\circ}-\theta)\). If \(\sin\theta = \cos\theta\), then \(\theta= 45^{\circ}\) because \(\sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}\) (since \(\cos(90^{\circ} - 45^{\circ})=\cos45^{\circ}=\sin45^{\circ}\)).

Step2: Analyze each triangle

• First triangle: Angles are \(40^{\circ},50^{\circ},90^{\circ}\). No \(45^{\circ}\) angle.
• Second triangle: Angles are \(45^{\circ},45^{\circ},90^{\circ}\) (isosceles right triangle). It has \(45^{\circ}\) angles, for which \(\sin45^{\circ}=\cos45^{\circ}\).
• Third triangle: Angles are \(22^{\circ},68^{\circ},90^{\circ}\). No \(45^{\circ}\) angle.
• Fourth triangle: Angles are \(30^{\circ},60^{\circ},90^{\circ}\). No \(45^{\circ}\) angle.

Answer:

The triangle with vertices \(A\), \(B\), \(C\) (the second triangle) where \(\angle A = 45^{\circ}\), \(\angle B=45^{\circ}\), \(\angle C = 90^{\circ}\) (the isosceles right - triangle with two \(45^{\circ}\) acute angles).