Find the perimeter of an isosceles triangle
whose base is 40 cm and whose base angle is 70".

$$cos70^o=\frac{20}{x}\\\\cos70^o\approx0.342\\\\\frac{20}{x}=0.342\\\\0.342x=20\ \ \ \ /:0.342\\\\x\approx58.5\ (cm)\\\\Perimeter:40cm+2\cdot58.5cm=157cm$$

$$cos70^0= \frac{20}{x}\\\\x= \frac{20}{cos70^0} \\\\the\ perimeter\ is:\\\\ P=2x+40=\frac{40}{cos70^0}+40 \ \ \ and\ \ \ cos70^0\approx0.3420\\\\P\approx\frac{40}{0.342}+40\approx156.96\ [cm]$$

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