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2 years ago

Use triangleABC to find the value of sin B

Mathematics

2 answers:

klasskru [66]2 years ago

7 0

The given triangle is a right triangle right angled at angle C. For angle B the adjacent side is BC ,the side opposite to angle B is AC while AB is hypotenuse. Sin of any angle is ratio of opposite and hypotenuse.

Sin $\frac{opposite}{hypotenuse}$

The opposite in the figure is 24 and hypotenuse is 25

Substituting the values of opposite and hypotenuse from figure we have:

Sin B= $\frac{24}{25}$

Delicious77 [7]2 years ago

4 0

As we are given

Opposite side of the angle B=24

Hypotenuse =25

$\\ \ sin \theta=\frac{Opposite}{Hypotenuse}\\ \\ \ SinB=\frac{24}{25}\\$

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Leona [35]

Answer:

Step-by-step explanation:

The question doesn't make sense. There is no way receipts could have totaled $8008.50. The most that could have been taken is if all 430 of the patrons were adults. That gives 430×2.25 = $967.50

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2 years ago

Find sin X, sin Z, cos X, and cos Z. Write each answer as a simplified fraction.

just olya [345]

Answer:

Find the answers below

Step-by-step explanation:

Using m<X as the reference angle

Opposite YZ = 7

Adjacent XY = 10

Hypotenuse XZ = √149

Using the SOH CAH TOA identity

sinX = opp/hyp

sinX =YZ/XZ

sinX = 7/√149

For cos X

cos X = adj/hyp

cos X =10/√149

Using m<Z as reference angle;

Opposite XY = 10

Adjacent YZ = 7

Hypotenuse XZ = √149

Using the SOH CAH TOA identity

sinZ = opp/hyp

sinZ =10/√149

sinZ = 7/√149

For cos Z

cosZ = 7/√149

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2 years ago

In this equation, what is n?

raketka [301]

N is the variable, you solve by subtracting 6 from 20, getting n = 14.

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3 years ago

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2 years ago

Harmonic mean if a number h such that (h-a)/(b-h)=a/b Prove h is H(a,b) iff satisfies either relation a. (1/a)-(1/h) = (1/h) - (

Fudgin [204]

A.

$\displaystyle\frac1a-\frac1h=\frac1h-\frac1b$
$\implies\displaystyle\frac{h-a}{ah}=\frac{b-h}{bh}$
$\implies\displaystyle\frac{h-a}{b-h}=\frac{ah}{bh}$
$\implies\displaystyle\frac{h-a}{b-h}=\frac ab$

b.

$\displaystyle h=\frac{2ab}{a+b}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{\frac{2ab}{a+b}-a}{b-\frac{2ab}{a+b}}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{2ab-a(a+b)}{b(a+b)-2ab}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{ab-a^2}{b^2-ab}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac{a(b-a)}{b(b-a)}$
$\displaystyle\implies\frac{h-a}{b-h}=\frac ab$

The other direction can be proved by following the manipulations in the reverse order.

7 0

2 years ago