The product is sin 30 and sin 60 is some as the product of and

To find the height of a pole, a surveyor moves 160 feet away from the base of the pole and then, with a transit 6 feet tall, mea

Artyom0805 [142]

Answer:

306.8 feet

Step-by-step explanation:

Refer the attached figure .

A surveyor moves 160 feet away from the base of the pole i.e. ED = 160 m

Referring the image .

AC=ED=160 m

A transit i.e. AE is 6 feet tall.

AE = CD=6 feet

With a transmit the angle of elevation to the top of the pole to be 62° i.e. ∠BAC = 62°

To Find BC , use tigonometric ratio.

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan 62^{\circ}=\frac{BC}{AC}$

$Tan 62^{\circ}=\frac{BC}{160}$

$1.88*160=BC$

$300.8=BC$

Now, the height of the pole = BC+CD =300.8+6=306.8 feet

Hence the height of the pole is 306.8 feet

7 0

3 years ago

Solutions of x2 + 13x + 30 = 0 by factoring.

VMariaS [17]

Answer:

x = -10 or x = -3

Step-by-step explanation:

Here, the given equation is

$x^{2} + 13x + 30 = 0$

The given equation can be solved by the method of SPLITTING THE MIDDLE TERM

Split 13 in such a way that sum of the terms = 13

and product of the terms = 30

So, the given equation becomes $x^{2} + 10x + 3x + 30 = 0$

Here, 10x+ 3x = 13x and 10 x 3 = 30

Now, simplifying the equation,

$x(x+10) + 3(x + 10) = 0 \\(x+3) (x+10) =0$

⇒ either (x+ 10) = 0 ,or (x+ 3) = 0

⇒ x = -10 or x = -3

4 0

3 years ago

What is the formula to find the length of an arc when given the central angle and radius?

vazorg [7]

$s = \dfrac{n}{360^\circ}2 \pi r$

where
s = arc length
n = measure of central angle
r = radius

3 0

3 years ago

What is the rate of change of the linear relationship modeled in the table?

Lisa [10]

The correct answer is 3/2.

Slope = Rate Of Change

Slope Formula: y₂ - y₁ 5 - 2 3
-------------- = ---------- = -----
x₂ - x₁ 3 - 1 2

I hope that this helps you!

Btw, ₁ and ₂ are subscripts, they do not have anything to do with the actual working of the problem.

7 0

3 years ago

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).

Therefore, AB must be exactly half of AD:

$AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24$

3 0

2 years ago

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