This question was uploaded on 24/08/21 on social media accounts.
Hypotenuse of a right angled triangle is divided in three parts by two lines starting from angle opposite to hypotenuse. Out of three parts, two parts are 9 and 6. Find angle between two lines dividing the hypotenuse.
Step 1:
Give names to all vertices and intersection of lines.
Step 2:
AC is hypotenuse.
\[\Rightarrow AB^2+BC^2=AC^2\]
\[\Rightarrow 15^2+8^2=AC^2\]
\[\Rightarrow AC=17\]
Given, AM and MN are 9 and 6 respectively.
⇒ CN = 17 - 9 - 6;
⇒ CN = 2;
Observe,
In ⧍ABN,
AB = AN;
⇒ ∠ABN = ∠ANB;
Let, ∠ABM = x;
⇒ ∠ABN = ∠ANB = x + θ;
Similarly,
In ⧍CBM,
CB = CM
⇒ ∠CBM = ∠CMB;
Let, ∠CBN = y
⇒ ∠CBM = ∠CMB = y + θ
Step 3:
Given,
∠ABC is right angle,
⇒ x + y + θ = 90° --------(1)
In ⧍BMN,
∠B + ∠M + ∠N = 180°
⇒ θ + (y + θ) + (x + θ) = 180°
⇒ 2θ + (x + y + θ) = 180°
⇒ 2θ + 90° = 180° --------(From (1))
⇒ 2θ = 90°
⇒ θ = 45°
Image Solution:
SOLUTION 2: (From Twitter)
Puzzle related to Geometry, Triangle, Angle.
