Question

find h and x. pls answer it now ​

Answer 1

Answer:

h = 5($\sqrt{3}$ - 1) m and x = 15 - 5$\sqrt{3}$ m

Step-by-step explanation:

tan 30° = $\frac{h}{x}$

∵ tan 30° = 1 / $\sqrt{3}$

1 / $\sqrt{3}$ = h / x

x = h$\sqrt{3}$ -------(1)

Also,

tan 45 = (h+10) / x

Since, tan 45 = 1

1 = (h + 10) / x

x = h + 10

From equation number (1), we get

h$\sqrt{3}$ = h + 10

10 = h$\sqrt{3}$ - h

10 = h($\sqrt{3}$ - 1)

Therefore,

h = 10 / ($\sqrt{3}$ - 1) m

Rationalizing the denominator, we get

h = 5($\sqrt{3}$ - 1) m

Now putting the value of h in equation number (1), we get

x = h$\sqrt{3}$

x = 5($\sqrt{3}$ - 1)($\sqrt{3}$)

x = 5(3 - $\sqrt{3}$) m.

Answer 2

Answer:

$\huge\boxed{x=5\left(3-\sqrt3\right);\ h=5\left(\sqrt3-1\right)}$

Step-by-step explanation:

Look at the picture.

From  traingle 45°, 45°, 90° we have:

(1) $x=h+10$

From triangle 30°, 60°, 90° we have:

(2) $x=h\sqrt3$

From (1) and (2) we have:

$h\sqrt3=h+10\\^{-h\qquad-h}\\h\sqrt3-h=10\\\\(\sqrt3-1)h=10\\^{:(\sqrt3-1)\qquad:(\sqrt3-1)}\\h=\dfrac{10}{\sqrt3-1}\cdot\dfrac{\sqrt3-1}{\sqrt3-1}\\\\h=\dfrac{10(\sqrt3-1)}{(\sqrt3)^2-1^2}\\\\h=\dfrac{10(\sqrt3-1)}{3-1}\\\\h=\dfrac{10(\sqrt3-1)}{2}\to \boxed{h=5(\sqrt3-1)}$

substitute it to (2)

$x=5(\sqrt3-1)\cdot\sqrt3=(5\sqrt3-5)\cdot\sqrt3=(5\sqrt3)(\sqrt3)-(5)(\sqrt3)=5\cdot3-5\sqrt3\\\\\boxed{x=5(3-\sqrt3)}$