If x = 45, y = 63, and the measure of AC = 4 units, what is the difference in length between segments AB and AD? Round your answer to the nearest hundredth.

The difference in the length of AB and AD is 1.17 units

Step-by-step explanation:

CHECK THE ATTACHMENT FOR FIGURE FOR THE QUESTION

If we study the figure in the attachment , In the ΔABC and ΔADC,

We were given;

x = 45°

y = 63°

AC = 4 units

We will be making use of Sine rule in ΔABC to get our AB side, and we know that in trigonometry, sine rule is an equation that relate the lengths of the sides of a triangle to the sines of its angles.

Sin(B)= Opposite/Hypotenuse

Sin(X) = AC/AB

Sin(45) = 4/AB

Then, Sin(45) * AB = 4

But Sin(45)=1/√2

Then if we substitute the value we have

5.657units

Hence, AB= 5.657units

We can also make use of by sine rule in ΔADC to get our AD side

Sin(y)° = AC/AD

Sin(63) = 4/AD

Sin(63)× AD= 4

AS= 4.489 units

Hence AD= 4.489 units

To calculate the difference in the length of AB and AD, we will need to substract side AD from AB; which is

AB - AD = 5.657 - 4.489

= 1.168

= 1.17 units ( if we approximate)

Hence, the difference in the length of AB and AD would be 1.17 units

$12. that is my answer.

1/2*qr*rs

step-by-step explanation: