What type of triangle is this?

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .

For triangle DEF:

The length of segment DF is to be found,
The length of segment EF is 9,
The sine of angle E is $\sin{64\textdegree}}$ , and
The sine of angle D is $\sin{39\textdegree}$ .

Apply the law of sine:

$\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}$

$\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9$ .

7 0

3 years ago

Eight plus the quotient of a number and 3<br> is −2<br><br> find answer

Fed [463]

Eight plus the quotient of a number and 3 is −2

8 + (x/3) = - 2

Hope it helps

8 0

3 years ago

How do I solve this?

maksim [4K]

Answer:

Answer A represents this

x <= -3 ; x > 9

Step-by-step explanation:

Isolate x on each of the inequalities:

4 x + 4 <= - 8

subtract 4 from both sides

4 x <= - 12

divide both sides by 4 to isolate x completely (notice 4 is positive, so there is NO flipping of the inequality symbol)

x <= -3

This inequality is represented by shading the left section of the number line starting at x = -3 (make sure you draw a SOLID dot to mae clear that the point x = -3 is also included in your drawing.

The other inequality:

11 x - 11 > 88

add 11 to both sides

11 x > 99

divide both sides by 11 to isolate x (notice again that 11 is a positive number, so the inequality doesn't change with the division)

x > 9

This inequality is represented by shadowing the right hand side of the number line starting at x = 9. Make sure you draw an EMPTY dot on the number 9 to indicate that x = 9 is NOT included.

4 0

3 years ago

Please help! Will mark brainliest! 20 points!

notsponge [240]

8 3

-

9 4

Get common denominators.

8(4) 3 (9)

-

9(4) 4 (9)

=

32 27

-

36 36

=

5/36

5 0

3 years ago

Which relationship describes angles 1 and 2?

Alexxx [7]

Conplementary angles

4 0

4 years ago

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