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slavikrds [6]
4 years ago
14

Help anymore this is in geometry

Mathematics
1 answer:
mojhsa [17]4 years ago
5 0

Answer:

\huge\boxed{x^o=80^o}

Step-by-step explanation:

<em>Look at the picture</em>.

We have:

l=160^o

Therefore

2\alpha=160^o         <em>divide both sides by 2</em>

\dfrac{2\alpha}{2}=\dfrac{160^o}{2}\\\\\alpha=80^o

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\large\underline{\sf{Solution-}}

<u>We need to solve,</u>

\dfrac{\sqrt{112}-\sqrt{80}}{\sqrt{20}-\sqrt{28}}

We can write the above mentioned expression as,

\sf\longmapsto\dfrac{\sqrt{2\times2\times2\times2\times7}-\sqrt{2\times2\times2\times5}}{\sqrt{2\times2\times5}-\sqrt{2\times2\times7}}

So,

\sf\longmapsto\dfrac{\sqrt{4\times4\times7}-\sqrt{4\times4\times5}}{\sqrt{2^2\times5}-\sqrt{2^2\times7}}

So,

\sf\longmapsto\dfrac{\sqrt{4^2\times7}-\sqrt{4^2\times5}}{\sqrt{2^2\times5}-\sqrt{2^2\times7}}

Hence,

\sf\longmapsto\dfrac{4\sqrt{7}-4\sqrt{5}}{2\sqrt{5}-2\sqrt{7}}

Taking common in respective terms,

\sf\longmapsto\dfrac{4(\sqrt{7}-\sqrt{5})}{2(\sqrt{5}-\sqrt{7})}

On cancelling 4 with 2,

\sf\longmapsto\dfrac{4\!\!\!/^{\:2}(\sqrt{7}-\sqrt{5})}{2\!\!\!/(\sqrt{5}-\sqrt{7})}

\sf\longmapsto\dfrac{2(\sqrt{7}-\sqrt{5})}{(\sqrt{5}-\sqrt{7})}

Taking (-) common,

\sf\longmapsto\dfrac{-2(\sqrt{5}-\sqrt{7})}{(\sqrt{5}-\sqrt{7})}

So, (√5 - √7) gets cut,

<u>Hence, </u>

\longmapsto\bf\dfrac{\sqrt{112}-\sqrt{80}}{\sqrt{20}-\sqrt{28}}=-2

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