This question was uploaded on 30/08/21 on social media accounts.
A circle is drawn inside a semicircle such that it will touch the center of the semicircle and it has a radius of half of that of the semicircle. Two chords are drawn inside the semicircle such that one chord starts from one end of diameter and ends on a common point of both circles. And other chord starts from another end of diameter and passes through the intersection of the small circle and the first chord. The third chord is drawn inside the small circle such that it starts and ends at end of the first chord and the intersection point of the second chord and the small circle. Find the tangent of the angle between the First and Third chords.
Solution:
Step 1:
Join center of circle and intersection points of chords of the large circle with a small circle.
Draw a perpendicular from intersection point the first chord of a large circle with a small circle on the diameter of the large circle.
Give names to all important points in this figure.
Let the radius of the small circle be 'r' and the radius of the large circle be '2r'.
Step 2:
Here angle POQ is twice of angle PRQ by inscribed angle theorem.
∠POQ = ∠PRQ = 2α
⇒ ∠OPQ = 90 - α
⇒ ∠BPN = α
Here P is the mid-point AR. This can be proved by triangles NAP and OPR. Both triangles are congruent.
⇒ AN = PO = NM = r;
MB = 2r;
Step 4:
In triangle BPN:
tan(α) = 3r / r = 3
Image Solution for easy Understanding:
Puzzle related with topics: Geometry, Circle, Angle.
