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A-piece-of-plastic-strip-1-metre-long-is-bent-to-form-an-isosceles-triangle-with-95-as-measure-of-its-largest-angle-Find-the-length-of-the-sides-




Question Number 5008 by Rasheed Soomro last updated on 31/Mar/16
A piece of plastic strip 1 metre long  is bent to form an isosceles triangle  with 95° as measure of its largest angle.  Find the length of the sides.
$$\mathrm{A}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{plastic}\:\mathrm{strip}\:\mathrm{1}\:\mathrm{metre}\:\mathrm{long} \\ $$$$\mathrm{is}\:\mathrm{bent}\:\mathrm{to}\:\mathrm{form}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\mathrm{95}°\:\mathrm{as}\:\mathrm{measure}\:\mathrm{of}\:\mathrm{its}\:\mathrm{largest}\:\mathrm{angle}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}. \\ $$
Commented by prakash jain last updated on 03/Apr/16
Angles 95°, 42.5°,42.5°  (a/(sin A))=(b/(sin B))=(c/(sin C))  b=c  a+2b=1  (a/(sin 95°))=(b/(sin 42.5°))  (a/(sin 95°))=((1−a)/(2sin 42.5°))
$$\mathrm{Angles}\:\mathrm{95}°,\:\mathrm{42}.\mathrm{5}°,\mathrm{42}.\mathrm{5}° \\ $$$$\frac{{a}}{\mathrm{sin}\:{A}}=\frac{{b}}{\mathrm{sin}\:{B}}=\frac{{c}}{\mathrm{sin}\:{C}} \\ $$$${b}={c} \\ $$$${a}+\mathrm{2}{b}=\mathrm{1} \\ $$$$\frac{{a}}{\mathrm{sin}\:\mathrm{95}°}=\frac{{b}}{\mathrm{sin}\:\mathrm{42}.\mathrm{5}°} \\ $$$$\frac{{a}}{\mathrm{sin}\:\mathrm{95}°}=\frac{\mathrm{1}−{a}}{\mathrm{2sin}\:\mathrm{42}.\mathrm{5}°} \\ $$

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