Ask question

Ask question

All categories

3 years ago

11

Triangle ABC pictured above has a right angle at C. If a=4.3 and c=9, find the measure of angle B. Round to the nearest tenth of

a degree.
I ONLY NEED THE ANSWER HELP MEEEEEE

1 answer:

Cerrena [4.2K]3 years ago

7 0

61.4596 degrees? I used law of sines

You might be interested in

melissa began atop a hill that was 1,350 feet high. she descended 120 feet, took a break and then descended 235 more feet. Which

Anna11 [10]

1500 is the answer I did this a while ago I think not sure

5 0

3 years ago

Read 2 more answers

Sachant que le volume de la glace diminue de 11% quand elle fond, donner une valeur approchée, au millimètre près, de la hauteur

igomit [66]

Only if I knew this language

8 0

3 years ago

svet-max [94.6K]

Answer:

The polygon has 9 sides

Step-by-step explanation:

we know that

A regular polygon has equal length sides and equal interior angles

step 1

Find the measure of the interior angle

Remember that

The interior angle plus the exterior angle must be equal to 180 degrees (form a linear pair)

Let

x ----> the measure of each interior angle in the regular polygon

so

$x+40^o=180^o$

solve for x

$x=180^o-40^o=140^o$

step 2

Find the number of sides of the regular polygon

we know that

The measure of each interior angle in a regular polygon is given by the formula

$x=\frac{(n-2)180^o}{n}$

where

n is the number of sides of the polygon

substitute the given values

$140^o=\frac{(n-2)180^o}{n}$

solve for n

$140n=180n-360\\180n-140n=360\\40n=360\\n=9\ sides$

therefore

The polygon has 9 sides

8 0

3 years ago

I need help with this question

Novay_Z [31]

Answer:

$\frac{\sqrt{3} - 1}{2\sqrt{2}}$

$\frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

$- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Step-by-step explanation:

Given $\frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6}$

(A) $sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4} + \frac{\pi}{6})$

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2}$

$$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

(B) Cos(A + B) = CosAcosB - SinASinB

$cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}})$

$\implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2}$

$\implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6})$

$$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

(C) Tan(A + B) = $\frac{Sin(A +B)}{Cos(A + B)}$

From the above obtained values this can be calculated and the value is $- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .

3 0

3 years ago

scoray [572]

Answer:

12

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers