In a right-angled triangle, the angle between the height and the bisector drawn from the apex

| September 3, 2021 | education

In a right-angled triangle, the angle between the height and the bisector drawn from the apex of the angle is 40 degrees. How to find the large angle?

Let ABC be a given right-angled triangle (angle C = 90 °). CH – height, CE – bisector (on the segment AB, points E and H lie in this sequence: A, E, H and B). ESN angle = 40 °.

Consider the CEN triangle: angle H = 90 ° (CH – height), angle C = 40 °, we calculate the value of the CEN angle: 180 ° – (90 ° + 40 °) = 180 ° – 130 ° = 50 °.

Consider a triangle CEB: angle E = 50 °, angle ECB is equal to half the angle ACB (since CE is a bisector). ECB angle = 90 °: 2 = 45 °. Let’s calculate the value of the angle B: 180 ° – (50 ° + 45 °) = 180 ° – 95 ° = 85 °.

In triangle ABC we calculate the value of angle A: 180 ° – (90 ° + 85 °) = 5 °.

Answer: the larger angle of the triangle (by cutting the right angle) is 85 °.



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