Advertisements
Advertisements
Question
In the figure, given below, ABCD is a square, and ∆ BEC is an equilateral triangle. Find, the case:∠ABE
Solution
We know that the sides of a square are equal and each angle is of 90°
Three sides of an equilateral triangle are equal and each angle is of 60.
Therefore, In fig., ABCD is a square and Δ BEC is an equilateral triangle.
(i) ∠ABE = ∠ABC + ∠CBE
= 90° + 60° = 150°
(ii) But in Δ ABE
∠ABE + ∠BEA + ∠BAE = 180° ...............(Angles of a triangle)
⇒ 150° + ∠BAE + ∠BAE = 180° .............(∵ AB = BE)
⇒ 150° + 2 ∠BAE = 180°
⇒ 2 ∠BAE = 180°− 150° = 30°
∴ ∠BAE =`(30°)/2=15°`
APPEARS IN
RELATED QUESTIONS
Find the unknown angles in the given figure:
Find the unknown angles in the given figure:
Find the unknown angles in the given figure:
The vertical angle of an isosceles triangle is 15° more than each of its base angles. Find each angle of the triangle.
The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle.
The ratio between a base angle and the vertical angle of an isosceles triangle is 1: 4. Find each angle of the triangle.
In the given figure, BI is the bisector of ∠ABC and Cl is the bisector of ∠ACB. Find ∠BIC.
In ∆ ABC, BA and BC are produced. Find the angles a and h. if AB = BC.
In Figure, BP bisects ∠ABC and AB = AC. Find x.
In the figure, AB is parallel to CD, find x