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AMBQID 64 ABC Right Isosceles Triangle BD = BD Find Theta

AMBQID 64:

In the given figure, the triangle ABC is a right isosceles triangle and BD = BC. Find the value of θ.

Options:

A. 22.5 Degree

A. 30 Degree

A. 7.5 Degree

D. 15 Degree

AMB QID 64

Solution:

To solve the above question, we need to use the following math concepts.

Concept 1: The perpendicular distance between the parallel lines is same at any points

Concept 2: In a right triangle, the Pythagoras theorem is used. It the square of the hypotenuse is the sum of the squares of the perpendicular and the square of the base. 

Concept 3: In a right isosceles triangle, the two perpendicular lines are equal and each acute angle is 45 degree.

Concept 4: In an isosceles triangle, the perpendicular drawn from the vertex to opposite side bisects the side.

Concept 5: In trigonometry, sin(Angle) = Perpendicular / Hypotenuse.

Let the sides of right isosceles triangle AB = AC = a. Then, BC = a\sqrt{2} by using Concept 2.

Since AB = AC = a, then angle ABC = 45 degree using concept 3.

Given that BD = BC. Thus BD = a\sqrt{2}.

Now, in isosceles triangle ABC, drawn a perpendicular AM from the vertex A to the opposite side BC. Thus, BM = MC = BC/2 by using concept 4. Hence, BM = \frac{a}{\sqrt{2}}.

In right triangle ABM, AM = \frac{a}{\sqrt{2}} using concept 2.

Since the lines AD and MN are parallel, then AM = DN by using concept 1. Thus, DN = \frac{a}{\sqrt{2}}. Now, let angle DBN = x.

Using concept 5, we get sin x = DN / BD = 1/2. Thus, x = 30 degree.

Therefore, θ = 45 degree – 30 degree = 15 degree. The correct option is D 15 degree.

 

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