Answer
Part A: The right triangles are smiliar because If the lengths of the hypotenuse and a leg of a right triangle are proportional to the corresponding parts of another right triangle, then the triangles are similar.
Part B: 2z:2x or 2z/2x. You can write z/x as z:x. Then just multiply both sides by the same number to get an equivalent ratio. Let's multiply by 2. then the answer will be: 2z:2x or 2z/2x
Step-by-step explanation:
For the considered two right triangles, the evaluations for sub-parts of the problem are as follows:
Two triangles are similar if two corresponding angles of them are of same measure. It is because when two pairs of angles are similar, then as the third angle is fixed if two angles are fixed, thus, third angle pair also gets proved to be of same measure. This makes all three angles same and thus, those two triangles are scaled copies of each other. Thus, they're called similar.
Trigonometric ratios for a right angled triangle are from the perspective of a particular non-right angle.
In a right angled triangle, two such angles are there which are not right angled(not of 90 degrees).
The slant side is called hypotenuse.
From the considered angle, the side opposite to it is called perpendicular, and the remaining side will be called base.
From that angle (suppose its measure is θ),
[tex]\sin(\theta) = \dfrac{\text{Length of perpendicular}}{\text{Length of Hypotenuse}}\\\cos(\theta) = \dfrac{\text{Length of Base }}{\text{Length of Hypotenuse}}\\\\\tan(\theta) = \dfrac{\text{Length of perpendicular}}{\text{Length of base}}\\\\\cot(\theta) = \dfrac{\text{Length of base}}{\text{Length of perpendicular}}\\\\\sec(\theta) = \dfrac{\text{Length of Hypotenuse}}{\text{Length of base}}\\\\\csc(\theta) = \dfrac{\text{Length of Hypotenuse}}{\text{Length of perpendicular}}\\[/tex]
For the considered case of the specified two right angled triangles, we get the solutions to the parts given in the problem as:
Part 1: The right angled triangles have one angles as of 90 degrees. In the image given, it can be seen that the other angle is of [tex]\alpha[/tex] measures, which is also common in both the triangles.
Thus, two angle pairs in both the given triangles are of same measure. That makes both the specified triangles as a pair of similar triangles.
Part 2: The specified ratio is z/x. From the perspective of angle [tex]\alpha[/tex] , we see that z is serving as hypotenuse, and x is base, thus, the ratio z/x is sec([tex]\alpha[/tex] ) (as sec of an angle is ratio of hypotenuse to base)
Part 3: From the perspective of angle [tex]\alpha[/tex] , the perpendicular is the side 'y' and the base is side with length 'x' (we're considering first triangle here, as we can consider any one of them as both are similar).
Thus, we get tan of [tex]\alpha[/tex] as:
[tex]\tan(\alpha) = \dfrac{y}{x}[/tex]
Part 4: The values of the trigonometric ratios depend only on the angle(as they're functions of angles only) because they are ratio.
If a triangle is right angled, and has a specific angle, then whatever the side be, if that angle along with right angle is kept same, the triangles will be similar . That makes their sides a specific multiple of each other.
a side of one triangle = k times corresponding side of its similar triangle
where k is the scale factor, constant for one pair of similar triangles. Now when we will take their ratio, the value of k cancels out, leaving no difference between first triangle or second triangle, but only caring if they are similar or not(which depends on angles).
Thus, the trigonometric functions depend only on angles.
Learn more about similar triangles here:
https://brainly.com/question/11929676
