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I know that this is a really easy question, but I am looking for the answer to this question:

The area of this isosceles triangle is 5cm squared.

The angle ABC is 22 degrees.

Work out the lengths of the two equal sides of the triangle

Thanks

Here is an image of the question

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    $\begingroup$ Which two sides are equal? $\endgroup$ – Kenny Lau Jun 25 '16 at 8:20
  • $\begingroup$ If $22^\circ$ is the included angle of those equal sides, then let $r$ be the equal sides, and then we have $\dfrac12r^2\sin(22^\circ)=5$. $\endgroup$ – Kenny Lau Jun 25 '16 at 8:20
  • $\begingroup$ can you post a picture please? $\endgroup$ – Dr. Sonnhard Graubner Jun 25 '16 at 8:21
  • $\begingroup$ Otherwise, the included angle of the equal sides would be $180^\circ-22^\circ-22^\circ=136^\circ$, so we would have $\dfrac12r^2\sin(136^\circ)=5$. $\endgroup$ – Kenny Lau Jun 25 '16 at 8:21
  • $\begingroup$ I have posted a photo $\endgroup$ – user350100 Jun 25 '16 at 8:23
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from $$5cm^2=\frac{1}{2}x^2\sin(22^{\circ})$$ we get $$x=\sqrt{\frac{10cm^2}{\sin(22^{\circ})}}$$ now you can take a calculator.

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  • $\begingroup$ I will try that now... $\endgroup$ – user350100 Jun 25 '16 at 8:29
  • $\begingroup$ So x=5.2? is that right? $\endgroup$ – user350100 Jun 25 '16 at 8:30
  • $\begingroup$ yes it is $$5.166688651887216543546753918323821954845281114471415318333742588216703336294167483471276050440784083$$ $\endgroup$ – Dr. Sonnhard Graubner Jun 25 '16 at 8:32

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