Question

Find the missing part.

Answer 1

Answer:

Answers are :

x = 7.5 , y = $\frac{15\sqrt{3} }{4}$ , z = $\frac{15\sqrt{3} }{2}$

a = 9.375 and b = 5.625

Step-by-step explanation:

From the attached figure , consider right triangle ABC.

∠B = 60°  , BC = 15  {∵ BC = a + b}

We need to find AC = z

Using sin function,

i.e sin(60°) = $\frac{AC}{BC}$

or sin(60°) = $\frac{AC}{15}$

or AC = 15×sin(60°)

or AC = $\frac{15\sqrt{3} }{2}$

Also, AB = x = 15×cos(60°) = $\frac{15}{2} = 7.5$

Next,

Consider right triangle ADC

AD = y, AC = z = $\frac{15\sqrt{3} }{2}$

∠C = 30°

Using sin function to get y.

i.e sin(30°) = $\frac{AD}{AC} = \frac{y}{z}$

or sin(30°) = \frac{y}{$\frac{15\sqrt{3} }{2}$}[/tex]

or y = $\frac{15\sqrt{3} }{2}$×sin(30°)

or y = $\frac{15\sqrt{3} }{4}$

Also, DC = b = $\frac{15\sqrt{3} }{2}$×cos(30°)

b = $\frac{45}{8} = 5.625$

Therefore, a= 15 - b = 15 - 5.625 = 9.375

Hence we got,

x = 7.5 , y = $\frac{15\sqrt{3} }{4}$ , z = $\frac{15\sqrt{3} }{2}$

a = 9.375 and b = 5.625

Answer 2

Answer:

$z = 15[\frac{\sqrt{3}}{2}]$

Step-by-step explanation:

To find the Z-side, we must take the cosine of the 30-degree angle of the main triangle

We know that the cosine of an angle is defined as:

$cos(30) = \frac{adjacent\ side}{hypotenuse}$

$cos(30) = \frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}} = \frac{z}{15}$

Then:

$z = 15[\frac{\sqrt{3}}{2}}]$

Finalmente the side z is:

$z = 15[\frac{\sqrt{3}}{2}}]$