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ludmilkaskok [199]
3 years ago
13

A right triangle with one angle that is 50*

Mathematics
1 answer:
denis-greek [22]3 years ago
7 0
Its an acute triangle
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Answer:

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

6(2 + √3)  

Step-by-step explanation:

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

To find: Radius<em>(KH)</em> =?

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

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}         (∵<em> tan 15°</em> = 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 <em>(KH)</em> = 6 (2+√3)

This is how to find the value of <em>tan 15°</em>

<em>tan</em> 15° = <em>tan </em>(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,

<em>tan</em> 15° = 2 - √3

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