Numeral succeeding 111 base 2

Which angles of the triangles measure 90°?

xxMikexx [17]

<EDA <ABC is da ones

6 0

3 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of

Llana [10]

Answer:

Part 1) $csc(\theta)=-\frac{\sqrt{85}}{7}$

Part 2) $sin(\theta)=-\frac{7}{\sqrt{85}}$ or $sin(\theta)=-\frac{7\sqrt{85}}{85}$

Part 3) $tan(\theta)=-\frac{7}{6}$

Part 4) $cos(\theta)=\frac{6}{\sqrt{85}}$ or $cos(\theta)=\frac{6\sqrt{85}}{85}$

Part 5) $sec(\theta)=\frac{\sqrt{85}}{6}$

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

so

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

$1+cot^{2}(\theta)=csc^{2}(\theta)$

we have

$cot(\theta)=-\frac{6}{7}$

substitute

$1+(-\frac{6}{7})^{2}=csc^{2}(\theta)$

$1+\frac{36}{49}=csc^{2}(\theta)$

$\frac{85}{49}=csc^{2}(\theta)$ rewrite

$csc(\theta)=-\frac{\sqrt{85}}{7}$ ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

$csc(\theta)=\frac{1}{sin(\theta)}$

we have

$csc(\theta)=-\frac{\sqrt{85}}{7}$

therefore

$sin(\theta)=-\frac{7}{\sqrt{85}}$

or

$sin(\theta)=-\frac{7\sqrt{85}}{85}$

step 3

Find the value of tan(θ)

we know that

$tan(\theta)=\frac{1}{cot(\theta)}$

we have

$cot(\theta)=-\frac{6}{7}$

therefore

$tan(\theta)=-\frac{7}{6}$

step 4

Find the value of cos(θ)

we know that

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$sin(\theta)=-\frac{7}{\sqrt{85}}$

substitute

$(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1$

$\frac{49}{85}+cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{49}{85}$

$cos^{2}(\theta)=\frac{36}{85}$

$cos(\theta)=\frac{6}{\sqrt{85}}$ ------> the cosine is positive

or

$cos(\theta)=\frac{6\sqrt{85}}{85}$

step 5

Find the value of sec(θ)

we know that

$sec(\theta)=\frac{1}{cos(\theta)}$

we have

$cos(\theta)=\frac{6}{\sqrt{85}}$

therefore

$sec(\theta)=\frac{\sqrt{85}}{6}$ ----> is positive

8 0

4 years ago

Please answer number 6

saw5 [17]

B should equal 10. you just have to figure out what would get you b and then you can figure it out from there!!

8 0

3 years ago

What would measure of angle APQ be? what would the measure of angle NPQ be?

Lorico [155]

The measure of angle APQ should be 90 degrees considering all the angles in a square add up to 360 degrees. NPQ should also be 90 degrees

6 0

3 years ago

PLEASE HELP ASAP!!!! Thanks!!

tino4ka555 [31]

Answer:

1.2 cm

Step-by-step explanation:

Quadrilateral circumscribing a circle is a quadrangle whose sides are tangent to a circle inside it (see attached diagram).

The area of circumscribed quadrilateral is

$A=p\cdot r,$

where $p=\dfrac{a+b+c+d}{2}$ is semi-perimeter and r is radius of inscribed circle.

In your case, $A=12\ cm^2$

If a quadrilateral is circumscribed over the circle, then the sum of opposite sides is equal, so

$a+c=b+d=10\ cm,$

so

$P=10+10=20\ cm\\ \\p=\dfrac{20}{2}=10\ cm$

Now

$12=10\cdot r\Rightarrow r=\dfrac{12}{10}=1.2\ cm$

8 0

4 years ago