The acute angle of a right-angled triangle is 38 degrees. Find the angle between the bisector and the height drawn from the vertex of the right angle
Let ABC be a right-angled triangle, angle A = 90 degrees, angle C = 38 degrees, AK – bisector, AH – height. Find angle B:
angle A + angle B + angle C = 180 degrees (according to the theorem on the sum of the angles of a triangle);
90 + angle B + 38 = 180;
angle B = 180 – 128;
angle B = 52 degrees.
1. Since AK is a bisector, it divides the angle A in half, then the angle KAВ = angle KAС = angle A / 2 = 90/2 = 45 degrees.
2. In the ВAK triangle, the angle AВK = angle B = 52 degrees, the angle KAВ = 45 degrees, AH is the height.
Let’s find the ВKA angle:
angle KAВ + angle ABK + angle ВKA = 180 degrees (according to the theorem on the sum of the angles of a triangle);
45 + 52 + angle ВКА = 180;
angle ВКА = 180 – 97;
angle ВКА = 83 degrees.
3. In the AНK triangle, the angle НKA = angle BCA = 83 degrees, the angle AНK = 90 degrees (since AH is the height, that is, the perpendicular dropped from the vertex A to the side BC).
Find the KAН angle:
KAN angle + AНK angle + НKA angle = 180 degrees (according to the theorem on the sum of the angles of a triangle);
KAN angle + 90 + 83 = 180;
KAN angle = 180 – 173;
KAN angle = 7 degrees.
Answer: KAN angle = 7 degrees.
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