Question

Solve each problem. (Source for Exercises 49 and 50: Parker, M., Editor, She Does Math, Mathematical Association of America.) Length of Sides of an Isosceles Triangle An isosceles triangle has a base of length 49.28 m. The angle opposite the base is 58.746°. Find the length of each of the two equal sides.

Accepted Answer

Identify the given elements of the isosceles triangle: the base length $$b = 49.28 m $$and the vertex angle opposite the base $$	heta = 58.746^\circ. $$The two equal sides are what we need to find. Recall that in an isosceles triangle, the two equal sides are opposite the equal angles. The vertex angle $$	heta is $$opposite the base, so the two equal sides meet at this vertex angle. Draw an altitude from the vertex angle to the base, which bisects the base into two equal segments of length $$\frac{b}{2} = \frac{49.28}{2} m $$and also bisects the vertex angle into two angles of $$\frac{	heta}{2} = \frac{58.746}{2}^\circ$$ each. Use the right triangle formed by the altitude, half the base, and one of the equal sides. Apply the cosine function to relate the half base and the equal side: $$\cos\left(\frac{	heta}{2}ight) = \frac{	ext{adjacent side}}{	ext{hypotenuse}} = \frac{\frac{b}{2}}{s}$$, where $$s is $$the length of each equal side. Rearrange the formula to solve for $$s$$: $$s = \frac{\frac{b}{2}}{\cos\left(\frac{	heta}{2}ight)}. $$Substitute the known values for $$b$$ and $$	heta to $$find the length of each equal side.

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주요 개념

Properties of Isosceles Triangles

Law of Cosines

Triangle Angle Sum Property