Question

Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB and sin ∠X = 5 over 5 and 59 hundredths. Triangles XYZ and ACB; angles Y and C both measure 90 degrees, angles A and X are congruent. Part A: Use complete sentences to explain the special relationship between the trigonometric ratios of triangles XYZ and ACB. You must show all work and calculations to receive full credit. (5 points) Part B: Explain how to find the measures of segments CB and AB. You must show all work and calculations to receive full credit. (5 points)

Answer 1

Answer:

See explanation

Step-by-step explanation:

Given

$\triangle XYZ \simeq \triangle ACB$

$\sin X = \frac{5}{5.59}$

$\angle Y = \angle C = 90^o$

$\angle A \cong \angle X$

$k = 2$ --- scale factor

See attachment for triangles

Solving (a): The relationship between the angles of both triangles

The sine of an angle is:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

By comparing the above to $\sin X = \frac{5}{5.59}$

$Opposite = 5\\ Hypotenuse = 5.59$

Using Pythagoras theorem:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

So, we have:

$5.59^2 = 5^2 + Adjacent^2$

$31.2481 = 25 + Adjacent^2$

Collect like terms

$31.2481 - 25 =Adjacent^2$

$6.2481 =Adjacent^2$

Take square roots

$Adjacent = 2.50$ --- approximated

For angle X, we have:

$Opposite = 5\\ Hypotenuse = 5.59\\ Adjacent = 2.50$

So, the relationship between both triangles are:

$\sin X =\sin A = \cos Z =\cos B= \frac{5}{5.59}$

$\cos X =\cos A =\sin Z =\sin B = \frac{2.50}{5.59}$

$\tan X =\tan A = \frac{5}{2.50} = 2$

$\tan Z =\tan B = \frac{2.50}{5} = \frac{1}{2}$

Solving (b): How to find CB and AB

In (a), we have:

For angle X, we have:

$Opposite = 5\\ Hypotenuse = 5.59\\ Adjacent = 2.50$

This implies that:

$XY = 2.50$

$YZ = 5$

$XZ = 5.59$

$\triangle XYZ$ is dilated by scale factor 2 to give $\triangle ACB$

So, CB and AB of $\triangle ACB$ are:

$CB = 2 * YZ$

$CB = 2 * 5$

$CB = 10$

$AB = 2 * XZ$

$AB = 2 * 5.59$

$AB = 11.18$