2.53 × 3.1 is between

Rectangle GHIJ inscribed in a circle, GK⊥JH, GK = 6cm, m∠GHJ = 15°. Find the radius of the circle. pls help !!

Mariulka [41]

Answer:

6(2 + √3)

Step-by-step explanation:

Given : Rectangle GHIJ inscribed in a circle. GK⊥JH, GK = 6 cm and m∠GHJ=15°.

To find: Radius(KH) =?

Sol: As given in figure 1, Since GK⊥JH ∴ m∠GKH = 90° . Let GK = x cm.

Now, In ΔGKH,

$tan\Theta =\frac{perpendicular}{base}$

$tan\Theta =\frac{GK}{KH}$

$tan 15^{\circ} =\frac{6}{x}$

$2-\sqrt{3} =\frac{6}{x}$ (∵ tan 15° = 2 - √3)

$x = \frac{6}{2-\sqrt{3} }$

On rationalizing the above expression,

$x = \frac{6}{2-\sqrt{3} } \times \frac{2+\sqrt{3} }{2+\sqrt{3} } =\frac{6(2+\sqrt{3} )}{4-3}$

Therefore, radius of the circle (KH) = 6 (2+√3)

This is how to find the value of tan 15°

tan 15° = tan (45° -30°)

Now using,

$tan(A-B) = \frac{tanA-tanB}{1+tanA tanB}$

$tan(45^{\circ} - 30^{\circ}) = \frac{tan 45^{\circ}-tan30^{\circ}}{1+tan 45 tan30}$

$tan(45^{\circ} - 30^{\circ}) = \frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

On rationalizing,

$tan 15^{\circ} = \frac{\left (\sqrt{3}-1 \right )^{2}}{3-1} =\frac{4-2\sqrt{3}}{2}$

Taking 2 common from numerator,

tan 15° = 2 - √3

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2 years ago

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How to use graphing to find the solutions to the system of equations?

gayaneshka [121]

Once ur lines are graphed, the solution is where the lines intersect. So by graphing however many equations u have, the point where they intersect is the solution....unless of course u have parallel lines, then there is no solution. And if ur lines are coincident, the same line on top of each other, then there is infinite solutions.

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The degree of the polynomial function f(x) is 3.

yawa3891 [41]

Answer:

3rd graph.

Step-by-step explanation:

Roots of the equation are where the graph crosses the x axis. So the bottom left (3rd) graph is correct.

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You row a canoe x miles per hour down a river for 20 minutes. On the return trip, you travel 1 mile per hour slower. The return

ozzi

Answer:

Total Distance = 2 miles

Step-by-step explanation:

We know D = RT where

D is distance

R is rate (speed)

T is time

First Leg:

Rate = x mph

Time = 20/60 = 1/3 hour

Hence, D = RT, Distance = x * 1/3 = x/3 miles

Second Leg:

Rate = x - 1

Time = 30/60 = 1/2 hour

Distance = (x - 1) * 1/2 = (x-1)/2

Total distance is the sum of both the legs, hence,

Total distance = $\frac{x}{3}+\frac{x-1}{2}=\frac{2x+3(x-1)}{6}=\frac{2x+3x-3}{6}=\frac{5x-3}{6}$ miles

Since both the distance are equal we can equate and solve:

$\frac{x}{3}=\frac{x-1}{2}\\2x=3x-3\\x=3$

Plugging this x = 3 into the total distance expression, we get:

$\frac{5x-3}{6}\\\frac{5(3)-3}{6}\\=2$

Total distance, 2 miles

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If you score 90% on an assignment, what is the ratio of correct questions to total questions?

artcher [175]

At least like...10,80 or 40

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