Find the value of x.

A 9m long ladder rests at 9m below a vertical wall. If the angle of elevation of the top of the wall from the foot of the ladder

Arte-miy333 [17]

Height of the wall is 27 m.

<u>Explanation:</u>

Let the height of the wall = h

Length of the ladder = 9 m

Angle of elevation = 60°

Ladder rests at 9 m below a vertical wall.

So, length of the wall upto which the ladder rests = h - 9

We know,

Cos Ф = base/ hypotenuse

cos 60° = length of the ladder / height of wall upto which the ladder rests

$\frac{1}{2} = \frac{9}{h - 9}$

So,

h - 9 = 18

h = 27

Therefore, height of the wall is 27 m

3 0

4 years ago

Between 02 30 and 05 15, the same day ,the volume of water in a reservoir increased from 1500 L to 14 700 L. What was the averag

marishachu [46]

Answer:

4,800

Step-by-step explanation:

(14,700- 1,500) / 2 h 45 m

= 13,200 / 2.75

= 4,800 L/h

6 0

3 years ago

Read 2 more answers

What are the coordinates of the vertex of the graph of

vesna_86 [32]

The vertex is (3,0). The 3 inside the abs val bar tells you the x value of the vertex is 3. No number outside the abs value bar is a +0 which is the y value of the vertex.

4 0

3 years ago

What is the measure of the missing angle x?

BARSIC [14]

Answer:

C. 60°

Step-by-step explanation:

Add the two given angles and subtract from 180°

4 0

3 years ago

Read 2 more answers

Find cos y and tan y if csc y = -√6/2 and cot y >0.

fomenos

Answer:

$\cos y = -\dfrac{\sqrt{3} }{3}$

$\tan y = \sqrt{2}$

Step-by-step explanation:

Recall that

$\boxed{\csc y := \dfrac{1}{\sin y}}$

$\boxed{\cot y := \dfrac{\cos y}{\sin y}}$

We know that

$\csc y = \dfrac{-\sqrt{6} }{2}$

Note that according to the definition of $\csc y$ it is true that both sine and cosine are negative, once $\csc y = \dfrac{-\sqrt{6} }{2}$ . Because $\cot y > 0$ , this conclusion is true. We basically have

$\boxed{(-a)(1/-b)=a/b \text{ such that } a,b\in\mathbb{R}_{\geq 0}}$

Sure it is true $\forall y\in\mathbb{R}$ but perhaps this way is better to understand.

In order to find sine, we can use the definition and manipulate the rational expression.

$\csc y = \dfrac{-\sqrt{6} }{2} = \dfrac{-\sqrt{6} / -\sqrt{6} }{2/-\sqrt{6} } = \dfrac{1 }{-\dfrac{2}{\sqrt{6} } }$

Therefore,

$\sin y =-\dfrac{2}{\sqrt{6} }$

Here I just divided numerator and denominator by $-\sqrt{6}$ .

Now, to find cosine we can use the identity

$\boxed{\sin^2y +\cos ^2y =1}$

Thus,

$\left(-\dfrac{2}{\sqrt{6} }\right)^2 + \cos ^2y =1 \implies \dfrac{4}{6 } +\cos ^2y =1$

$\implies \cos ^2y =1 - \dfrac{4}{6 } \implies \cos ^2y =\dfrac{1}{3 } \implies \cos y = \pm \dfrac{\sqrt{1} }{\sqrt{3} } = \pm \dfrac{\sqrt{1} \sqrt{3} }{3} = \pm \dfrac{\sqrt{3} }{3}$

$\cos y = \pm\dfrac{\sqrt{3} }{3}$

Once we have $\cot y > 0$ , we just consider

$\cos y = -\dfrac{\sqrt{3} }{3}$

FInally, for tangent, just consider

$\boxed{\tan y := \dfrac{\sin y}{\cos y}}$

thus,

$\tan y = \dfrac{\sin y}{\cos y} = \dfrac{-\frac{2}{\sqrt{6} }}{-\frac{\sqrt{3} }{3}} = \dfrac{6}{\sqrt{18} } =\dfrac{6}{3\sqrt{2} } =\dfrac{2}{\sqrt{2} } = \sqrt{2}$

5 0

3 years ago