Question

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

Answer 1

Answer:

Radius of circle is given as 22.39 cm

Step-by-step explanation:

Given data:

GHIJ is rectangle which is inscribe in a circle such that prependicular lie on one side of traingle form by joining the diagnol of rectangle is 6 cm

and angle is given as 15 degree as shown in figure given below

we know that

$tan\theta = \frac{prependicular}{ base}$

so we have

$tan\theta = \frac{6}{x}$

$tan15 = \frac{6}{x}$

tan 15  = 0.267 {from calculator}

so we have

x = 22.39 cm

hence the radius of circle is given as 22.39 cm

Answer 2

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