All categories

3 years ago

Find the value of tan theta if sin theta = 12/13 and theta is in quadrant 2

Mathematics

1 answer:

ivanzaharov [21]3 years ago

3 0

Answer:

اhello : tan θ = - 12/5

Step-by-step explanation:

tan θ = sin θ / cos θ .... (*)

(cosθ)² + (sinθ)² = 1 ... (**)

theta is in quadrant 2 : cosθ ≤ 0

Substitute sinθ = 12/13 into (**) and solve for cosθ :

(cosθ)² + (12/13)² = 1

(cosθ)² = 1 - 144/169

(cosθ)² = 25/169

cosθ = - 5 /13 because cosθ ≤ 0

by (*) : tan θ = (12/13)/ (-5/13) = (12/13) ×(-13/5)

tan θ = - 12/5

You might be interested in

Elimination method: ax+by=r, -bx+cy=s

Alex17521 [72]

$\left[\begin{array}{ccc}ax+by=r\\\\-bx+cy=s\end{array}\right]$

Isolate x for $ax+by=r$

$x= \frac{r-by}{a} ; a \neq 0$

Subsititute $x= \frac{r-by}{a}$

$-b \frac{r-by}{a} +cy=s$

Isolate r for : $-b \frac{r-by}{a}+cy=s$

$r=- \frac{as-acy-b^2y}{b} ; a \neq 0;b \neq 0$

For x : $\frac{r-by}{a}$

Subsititute r = $- \frac{as-acy-b^2y}{b}$

$x= -\frac{as-acy-b^2y}{b}/a$

$x = \frac{cy-s}{b}$

The solutions to the system of equations are:

$r= -\frac{as-acy-b^2y}{b}$ and <span> $x = \frac{cy-s}{b}$
</span>
hope this helps!

3 0

2 years ago

What is the solution to this equation?

sattari [20]

Answer:

Step-by-step explanation:

x - 9 = 17

x = 26

the answer is D

4 0

3 years ago

find the volume of a cylinder with a diameter of 10 inches and a height that is 3 times the radius. Use the 3.14 for pie and rou

Paul [167]

Answer:

1177.5 inches

Step-by-step explanation:

Volume of a cylinder = $\pi r^{2} *h$ , where h is the height

Radius = 10/2 = 5 radius is half the diameter

Height = 3*5 = 15 height is 3 times the radius

V = 3.14 * (5^2) * (15)

= 3.14 *25 * 15

= 1177.5

6 0

3 years ago

Consider the function f(x) = 2^x

Sonja [21]

Answer:

The graph of function g is the graph of function f shifted 6 units up

Step-by-step explanation:

If you plug in the values, $g(x) = 2^{x} + 6$ . If the 6 was added or subtracted from the x in the exponent, it would shift horizontally (left and right), but adding 6 to f(x) separately moves the graph vertically (up and down). Hope this helps.

4 0

3 years ago

Given the vectors A⃗ and B⃗ shown in the figure ((Figure 1) ), determine the magnitude of B⃗ −A⃗. A is 28 degrees above the posi

Vlad [161]

This problem is represented in the Figure below. So, we can find the components of each vector as follows:

$\bullet \ cos(28^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{A_{x}}{44} \\ \\ \therefore A_{x}=44cos(28^{\circ})=38.85m \\ \\ \\ \bullet \ sin(28^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{A_{y}}{44} \\ \\ \therefore A_{y}=44sin(28^{\circ})=20.65m$

$\bullet \ cos(56^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{-B_{x}}{26.5} \\ \\ \therefore B_{x}=-26.5cos(56^{\circ})=-14.81m \\ \\ \\ \bullet \ sin(56^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{B_{y}}{26.5} \\ \\ \therefore B_{y}=26.5sin(56^{\circ})=21.97m$

Therefore:

$\vec{A}=(38.85, 20.65)m \\ \\ \vec{B}=(-14.81, 21.97)m$

So:

$\vec{B}-\vec{A}=(-14.81, 21.97)-(38.85, 20.65)=(-53.66,1.32)$

Finally, the magnitude is:

$\boxed{\left| \vec{B}-\vec{A}\right|=\sqrt{(-53.66)^2+(1.32)^2}=53.67m}$

7 0

3 years ago

Read 2 more answers