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2 years ago

Use the figure to find the trigonometric ratio below. Express the answer as a decimal rounded to the nearest ten-thousandth.

Mathematics

2 answers:

Ira Lisetskai [31]2 years ago

8 0

Answer:

The correct option is 1.

Step-by-step explanation:

Given information: AD = 25, CD = 5, DB = 1 and CD⊥AB.

According to the Pythagoras theorem,

$hypotenuse^2=base^2+perpendicular^2$

In triangle BCD,

$CB^2=DB^2+CD^2$

$CB^2=1^2+5^2$

$CB^2=26$

Taking square root both sides.

$CB=\sqrt{26}$

In a right angled triangle,

$\sin\theta=\frac{opposite}{hypotenuse}$

$\sin B=\frac{CD}{CB}$

$\sin B=\frac{5}{\sqrt{26}}$

$\sin B=0.980580675691$

$\sin B\approx 0.9806$

Therefore the correct option is 1.

OlgaM077 [116]2 years ago

6 0

Answer:

0.9806 is the correct answer.

Step-by-step explanation:

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11. Given: TQ bisects ZRTP, QS || PT RT = 30. Find QP, RS and QS

yaroslaw [1]

QP=24 cm

RS=11.25 cm

QS=18.75 cm

Explanation:

Given that TQ bisects <RTP

$therefore$ (1)

consider ΔRQS and ΔRPT

QS||PT,RP and RT are transversals

(alternate angles)(2)

comparing (1) and (2)

and triangle SQT is isocelus

Therefore SQ=ST(sides opposite to equal angles in an isocelus triangle)

Therefore <RQS=<RPT(corresponding angles)

<RSQ=<RTP(corresponding angles)

therefore by AA criterion for similarity

ΔRQS~ΔRPT

According to the property of similar triangles

$RQ/RP=RS/RT=QS/PT$ $9/RP=X/30=QS/50\\9/RP=X/30=(30-X)/50\\X/30=(30-X)/50\\50X=30(30-X)\\50X=900-30X\\50X+30X=900\\80X=900\\X=900/80=11.25\\RS=11.25 cm\\QS=30-X=30-11.25\\QS=18.75 cm\\9/RP=X/30\\9/RP=11.25/30\\9*30/11.25=RP\\RP=24 cm$

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3 years ago

What two numbers add to 4 and multiply to 1

Arlecino [84]

There are no numbers for this, thus you will have to use the quadratic formula.

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2 years ago

Veronica and Chad are walking down the street. Veronica walks at a speed of 5 miles per hour, and Chad walks at a speed of 4 mil

stich3 [128]

Answer:

Veronica:

d=5t

Chad:

d=4t

Step-by-step explanation

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99 POINT QUESTION, PLUS BRAINLIEST!!!

sattari [20]

We know, that the area of the surface generated by revolving the curve y about the x-axis is given by:

$\boxed{A=2\pi\cdot\int\limits_a^by\sqrt{1+\left(y'\right)^2}\, dx}$

In this case a = 0, b = 15, $y=\dfrac{x^3}{15}$ and:

$y'=\left(\dfrac{x^3}{15}\right)'=\dfrac{3x^2}{15}=\boxed{\dfrac{x^2}{5}}$

So there will be:

$A=2\pi\cdot\int\limits_0^{15}\dfrac{x^3}{15}\cdot\sqrt{1+\left(\dfrac{x^2}{5}\right)^2}\, dx=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\left(\star\right)\\\\-------------------------------\\\\
\int x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\,dx=\int\sqrt{1+\dfrac{x^4}{25}}\cdot x^3\,dx=\left|\begin{array}{c}t=1+\dfrac{x^4}{25}\\\\dt=\dfrac{4x^3}{25}\,dx\\\\\dfrac{25}{4}\,dt=x^3\,dx\end{array}\right|=\\\\\\$

$=\int\sqrt{t}\cdot\dfrac{25}{4}\,dt=\dfrac{25}{4}\int\sqrt{t}\,dt=\dfrac{25}{4}\int t^\frac{1}{2}\,dt=\dfrac{25}{4}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}= \dfrac{25}{4}\cdot\dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}=\\\\\\=\dfrac{25\cdot2}{4\cdot3}\,t^\frac{3}{2}=\boxed{\dfrac{25}{6}\,\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}}\\\\-------------------------------\\\\$

$\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\=
\dfrac{5\pi}{9}\left[\left(1+\dfrac{15^4}{25}\right)^\frac{3}{2}-\left(1+\dfrac{0^4}{25}\right)^\frac{3}{2}\right]=\dfrac{5\pi}{9}\left[2026^\frac{3}{2}-1^\frac{3}{2}\right]=\\\\\\=
\boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}$

Answer C.

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Answer:

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Step-by-step explanation:

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