7 - (c + 18) > 0
-1 (c + [11])
~Now multiply each of the sides by -1~
c + 11 < 0
~Then , subtract 11 from each of the sides that you're working with~
c < -11
~Lastly, "c" becomes a negative~
-c - 11.0 > 0

Therefore your answer would have to be:
c < -11

8 0

3 years ago

Use the law of Sines to find the missing angle of the triangle. <br> Find m

Whitepunk [10]

Ah...Trigonometry is fun!

The law of sines states:
$\frac{sinA}{a}= \frac{sinB}{b} = \frac{sinC}{c}$
The transitive property (switching the orders of the equations) applies here. Therfore, we can say that $\frac{sinA}{a}=\frac{sinC}{c}$

We then plug in our given values to find C

$\frac{sin24}{77}=\frac{sinC}{162}$
$\frac{162*sin24}{77}=sinC$
Solving, we get 0.8557316387. We're not done yet!

We are trying to find an angle measure, so we'll do the inverse of the ratio we used (sin).

arcsin0.8557316387 (arcsin is the same as inverse sin)
=58.8 (approximate)

So the measure of angle C is 58.8. You could check this by reinserting it into the equation $\frac{sinA}{a}= \frac{sinB}{b} = \frac{sinC}{c}$ .

:)

6 0

3 years ago

Help please it's number lines

skelet666 [1.2K]

sin(θ+β)=−

−4

Step-by-step explanation:

step 1

Find the sin(\theta)sin(θ)

we know that

Applying the trigonometric identity

sin^2(\theta)+ cos^2(\theta)=1sin

(θ)+cos

(θ)=1

we have

cos(\theta)=-\frac{\sqrt{2}}{3}cos(θ)=−

substitute

sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1sin

(θ)+(−

)

sin^2(\theta)+ \frac{2}{9}=1sin

(θ)+

sin^2(\theta)=1- \frac{2}{9}sin

(θ)=1−

sin^2(\theta)= \frac{7}{9}sin

(θ)=

sin(\theta)=\pm\frac{\sqrt{7}}{3}sin(θ)=±

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

sin(\theta)=-\frac{\sqrt{7}}{3}sin(θ)=−

step 2

Find the sec(β)

Applying the trigonometric identity

tan^2(\beta)+1= sec^2(\beta)tan

(β)+1=sec

(β)

we have

tan(\beta)=\frac{4}{3}tan(β)=

substitute

(\frac{4}{3})^2+1= sec^2(\beta)(

)

+1=sec

(β)

\frac{16}{9}+1= sec^2(\beta)

+1=sec

(β)

sec^2(\beta)=\frac{25}{9}sec

(β)=

sec(\beta)=\pm\frac{5}{3}sec(β)=±

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

sec(\beta)=\frac{5}{3}sec(β)=

Remember that

sec(\beta)=\frac{1}{cos(\beta)}sec(β)=

cos(β)

therefore

cos(\beta)=\frac{3}{5}cos(β)=

step 3

Find the sin(β)

we know that

tan(\beta)=\frac{sin(\beta)}{cos(\beta)}tan(β)=

cos(β)

sin(β)

we have

tan(\beta)=\frac{4}{3}tan(β)=

cos(\beta)=\frac{3}{5}cos(β)=

substitute

(4/3)=\frac{sin(\beta)}{(3/5)}(4/3)=

(3/5)

sin(β)

therefore

sin(\beta)=\frac{4}{5}sin(β)=

step 4

Find sin(θ+β)

we know that

sin(A + B) = sin A cos B + cos A sin Bsin(A+B)=sinAcosB+cosAsinB

In this problem

sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)sin(θ+β)=sin(θ)cos(β)+cos(θ)sin(β)

we have

sin(\theta)=-\frac{\sqrt{7}}{3}sin(θ)=−

cos(\theta)=-\frac{\sqrt{2}}{3}cos(θ)=−

sin(\beta)=\frac{4}{5}sin(β)=

cos(\beta)=\frac{3}{5}cos(β)=

substitute the given values in the formula

sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})sin(θ+β)=(−

)(

)+(−

)(

)

sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})sin(θ+β)=(−3

)+(−4

)

sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}sin(θ+β)=−

−4

Step-by-step explanation:

i hope it helps to you

5 0

3 years ago