Help me to find AC please

Mathematics

1 answer:

spin [16.1K]2 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

Since the triangle is right use cosine or sine to solve for AC

note cos30° = $\frac{\sqrt{3} }{2}$ , thus

cos30° = $\frac{adjacent}{hypotenuse}$ = $\frac{AC}{BC}$ = $\frac{AC}{\frac{10}{3} }$ and

$\frac{AC}{\frac{10}{3} }$ = $\frac{\sqrt{3} }{2}$ , that is

$\frac{3x}{10}$ = $\frac{\sqrt{3} }{2}$ ( cross- multiply )

6x = 10 $\sqrt{3}$ ( divide both sides by 6 )

x = AC = $\frac{10}{6}$ $\sqrt{3}$ = $\frac{5}{3}$ $\sqrt{3}$

You might be interested in

Let c be parametrized by x = et sin (10t) and y = et cos (10t) for 0 ≤ t ≤ 2. find the length l of

marysya [2.9K]

$C:x(t)=e^t\sin10t;\,y(t)=e^t\cos10t;\,0\le t\le2$

$\displaystyle\int_0^2\sqrt{x'(t)^2+y'(t)^2}\,\mathrm dt$
$\displaystyle=\int_0^2\sqrt{(e^t\sin10t+10e^t\cos10t)^2+(e^t\cos10t-10e^t\sin10t)^2}\,\mathrm dt$
$\displaystyle=\int_0^2e^t\sqrt{\sin^210t+20\sin10t\cos10t+100\cos^210t+\cos^210t-20\sin10t\cos10t+100\sin^210t}\,\mathrm dt$
$=\displaystyle\int_0^2e^t\sqrt{1+100}\,\mathrm dt$
$=\displaystyle\sqrt{101}(e^2-1)$

7 0

3 years ago

Can anyone answer this?

Slav-nsk [51]

Answer:

601

Step-by-step explanation:

real real estate agents for

3 0

3 years ago

The number of phone calls between two cities, N, during a given time period varies directly as the populations p 1 and p 2 of t

kherson [118]