Answer:
The correct option: (2) Triangle ABC that has angle measures 45°, 45° and 90°.
Step-by-step explanation:
It is provided that a triangle ABC has an acute angle for which the sine and cosine ratios are equal to 1.
Let the acute angle be m∠A.
For the sine and cosine ratio of m∠A to be equal to 1, the value of Sine of m∠A should be same as value of Cosine of m∠A.
The above predicament is possible for only one acute angle, i.e. 45°, since the value of Sin 45° and Cos 45° is,
[tex]Sin\ 45^{o} =Cos\ 45^{o} = \frac{1}{\sqrt{2} }[/tex]
So for acute angle 45° the ratio of Sin 45° and Cos 45° is:
[tex]\frac{Sin\ 45^{o}}{Cos\ 45^{o}} = \frac{\frac{1}{\sqrt{2} } }{\frac{1}{\sqrt{2} } } = 1[/tex]
Hence one of the angles of a triangle is, m∠A = 45°.
Comparing with the options provided the triangle is,
Triangle ABC that has angle measures 45°, 45° and 90°.
Thus, the provided triangle is a right angled isosceles triangle, since it has two similar angles.
