The angle at the base of an isosceles triangle ABC (AB = BC) is 15 °, and the lateral side is 8. Find the distance from the vertex C to line AB.

| January 13, 2021 | education

Let the point of intersection of AB with the straight line from vertex C to line AB be point K.
And the point at which the height to AC from vertex B intersects AC will be D.
Consider a triangle ABD. Since BD is the height in ABC, therefore, it forms a right angle with AC, that is, ABD is a right-angled triangle. We know the length of the hypotenuse AB = 8 and the angle at the leg AD – 15º. Find AD:
AD = cos15º * 8 = √ (2 + √3) / 2 * 8 = 7.73.
Now consider the triangle AKC. KС is the minimum distance from C to AB, which means KС is perpendicular to AB. The AKC triangle is also rectangular, with an AC hypotenuse and an angle against the KС leg – 15º.
AC = AD * 2 = 7.73 * 2 = 15.46.
КС = sin15º * 15.46 = √ (2 – √3) / 2 * 15.46 = 4.
Answer: distance from vertex C to line AB 4.



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