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.

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

eudora eudora · Answer 2 · 2022-01-20T09:32:19+01:00

Given question is incomplete without the figure; find the figure attached.

Difference in the lengths of AB and AD is 1.17 units.

Given in the question,

m∠x = 45° and m∠y = 63°
AC = 4 units

Apply sine rule in ΔABC,

sin(x°) = [tex]\frac{AC}{AB}[/tex]

sin(45°) = [tex]\frac{4}{AB}[/tex]

AB = 4√2

AB ≈ 5.657

Similarly, apply sine rule in ΔADC,

sin(y°) = [tex]\frac{AC}{DC}[/tex]

sin(63°) = [tex]\frac{4}{AD}[/tex]

AD = 4.489

Now difference in lengths of AB and AD = 5.657 - 4.489

= 1.168

≈ 1.17

Therefore, difference in the lengths of AB and AD will be 1.17 units.

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