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Daniel [21]
3 years ago
5

How do I find Trig ratios

Mathematics
1 answer:
nadya68 [22]3 years ago
8 0

Trig ratios can only be used on right triangles with acute measures.

If given an angle and there are adjacent and opposite sides, then use tan(opposite/adjacent)

If given an angle and there is an adjacent side and a hypotenuse, then use cosine(adjacent/hypotenuse)

If given an angle and there is an opposite and adjacent side, then use sin(opposite/hypotenuse)

A common mnemonic device used to memorize the trig rules is SOH-CAH-TOA

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A bird (B) is spotted flying 6,000 feet from a tower (T)). An observer (O) spots the top of the tower (T) at a distance of 9,000
Nina [5.8K]

Answer:

The angle will be "48.7°".

Step-by-step explanation:

In ΔTBO, the given values are:

TB = 6,000 feet

BO = 9,000 feet

Now,

To find the angle,

⇒  Cos \ B =\frac{TB}{BO}

On putting the estimated values, we get

⇒   Cos \ x=\frac{6000}{9000}

⇒   Cos \ x=0.66

⇒          x=Cos^{-1} \ 0.66

⇒          x=48.7^{\circ}    

7 0
3 years ago
Read 2 more answers
X^3+3x^5=x5 what's the missing expression
shutvik [7]
<span><span><span>x3</span>+3x5</span>=x5</span><span>3x5+x3=x5</span>Subtract x^5 from both sides.<span>3x5+x3−<span>x5</span>=x5−<span>x5</span></span><span>2x5+x3=0</span>Factor left side of equation.<span><span><span>x3</span>(2x2+1)</span>=0</span>Set factors equal to 0.<span><span><span>x3</span>=0‌‌ or ‌‌2x2+1</span>=0</span><span>x=<span>0</span></span>
3 0
3 years ago
Read 2 more answers
Which relation is a function?
IceJOKER [234]

Answer:

The bottom right option

Step-by-step explanation:

Functions need each input to map to exactly one output. Graphically speaking, this means that any two points on the function cannot have the same x-coordinate.

Given this, the bottom right option is the only one that fits these circumstances.

4 0
3 years ago
PLEASE HELP IM ON A TIME LIMIT :(
Mice21 [21]

Answer:

hey hope this helps

<h3 /><h3>Comparing sides AB and DE </h3>

AB =

\sqrt{ {1}^{2} +  {1}^{2}  }

=  \sqrt{2}

DE

= \sqrt{ {(3 - 5)}^{2} +  {(1 + 1}^{2}  }  \\   = \sqrt{ {( - 2)}^{2}  +  {(2)}^{2} }  \\    = \sqrt{4 + 4}  \\  =  \sqrt{8}  \\   = 2 \sqrt{2}

So DE = 2 × AB

and since the new triangle formed is similar to the original one, their side ratio will be same for all sides.

<u>scale factor</u> = AB/DE

= 2

It's been reflected across the Y-axis

<em>moved thru the translation of 3 units towards the right of positive x- axis </em>

for this let's compare the location of points B and D

For both the y coordinate is same while the x coordinate of B is 0 and that of D is 3

so the triangle has been shifted by 3 units across the positive x axis

7 0
3 years ago
Enter an equation for the function that includes the points.Give your answer in a(b)x. In the event that a=1 , give your answer
Andrews [41]

Answer:

f(x) = \frac{24}{25} * \frac{5}{6}^x

Step-by-step explanation:

Given

(x_1,y_1) = (2,\frac{2}{3})

(x_2,y_2) = (3,\frac{5}{9})

Required

Write the equation of the function f(x) = ab^x

Express the function as:

y = ab^x

In: (x_1,y_1) = (2,\frac{2}{3})

y = ab^x

\frac{2}{3} = a * b^2 --- (1)

In (x_2,y_2) = (3,\frac{5}{9})

y = ab^x

\frac{5}{9} = a * b^3 --- (2)

Divide (2) by (1)

\frac{5}{9}/\frac{2}{3} = \frac{a*b^3}{a*b^2}

\frac{5}{9}/\frac{2}{3} = b

\frac{5}{9}*\frac{3}{2} = b

\frac{5}{3}*\frac{1}{2} = b

\frac{5}{6} = b

b = \frac{5}{6}

Substitute 5/6 for b in (1)

\frac{2}{3} = a * b^2

\frac{2}{3} = a * \frac{5}{6}^2

\frac{2}{3} = a * \frac{25}{36}

a = \frac{2}{3} * \frac{36}{25}

a = \frac{2}{1} * \frac{12}{25}

a = \frac{24}{25}

The function: f(x) = ab^x

f(x) = \frac{24}{25} * \frac{5}{6}^x

7 0
3 years ago
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