In triangle ABC, if m∠B = 90°, BH = AH, and the ratio of m∠A to m∠C is 1:2, then m∠BHA = 120° (C).
In the next given equilateral triangle ABC, y = 70°.
Finding m∠BHA:
In triangle ABC, it is given that,
∠B = 90°
And m∠A : m∠C = 1:2
Let us assume m∠A is x. Then, m∠C = 2x
According to the angle sum property of a triangle,
∠A + ∠B + ∠C = 180°
90° + x + 2x = 180°
90° + 3x = 180°
3x = 90°
x = 30°
⇒ In triangle ABC, ∠A = 30° and ∠C = 60°
Now, in triangle AHB, it is also given that,
BH = AH
⇒ ∠ABH = ∠A = 30°
Thus, according to the angle sum property of a triangle,
∠ABH + ∠A + ∠BHA = 180°
30° + 30° + ∠BHA = 180°
∠BHA = 180° - 60°
∠BHA = 120°
Finding y in the Second Triangle:
Since triangle ABC is equilateral,
∠A = ∠B = ∠C = 60°
∴ x + 2x + 3x = 60°
6x = 60°
x = 10°
In triangle ABD, using angle sum property of triangle,
x + ∠B + ∠BDA = 180°
10° + 60° + ∠BDA = 180°
∠BDA = 180° - 70°
∠BDA = 110°
Now, since, ∠BDA and y are linearly adjacent angles,
∠BDA + y = 180°
110° + y = 180°
y = 180° - 110°
y = 70°
Learn more about a triangle here:
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