Question : In equilateral $\triangle $ABC, D and E are points on the sides AB and AC, respectively, such that AD = CE. BE and CD intersect at F. The measure (in degrees) of $\angle $CFB is:
Option 1: 120°
Option 2: 135°
Option 3: 125°
Option 4: 105°
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 120°
Solution :
In $\triangle CBE$ and $ \triangle ACD$ AD = CE (Given) $\angle $B = $ \angle $C (each angle is 60º) BC = AC (sides of an equilateral triangle) $\triangle CBE \cong \triangle ACD$ [SAS congruency] So, the three angles of these two triangles are the same, Let $\angle $EBC = z then $\angle $ACD = z Now, $\angle $BEC = 180° – (60° + z) = 120° – z In $\triangle $ECF $\angle $CFB = (120° – z) + z (The exterior angle is always equal to the sum of the interior opposite angle) $\angle $CFB = 120° Hence, the correct answer is 120°.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : ABC is an isosceles triangle such that AB = AC, $\angle$ABC = 55°, and AD is the median to the base BC. Find the measure of $\angle$BAD.
Question : D and E are two points on the sides AC and BC, respectively of $\triangle ABC$ such that DE = 18 cm, CE = 5 cm, and $\angle$DEC = 90º. If $ \tan\angle$ABC = 3.6, then AC : CD = ?
Question : $D$ and $E$ are points on the sides $AB$ and $AC$ respectively of $\triangle ABC$ such that $DE$ is parallel to $BC$ and $AD: DB = 4:5$, $CD$ and $BE$ intersect each other at $F$. Find the ratio of the areas of $\triangle DEF$ and $\triangle CBF$.
Question : In an isosceles $\triangle ABC$, $AB = AC$, $XY || BC$. If $\angle A=30°$, then $\angle BXY$?
Question : ABC is a right angle triangle and $\angle ABC = 90^{\circ}$. BD is perpendicular on the side AC. What is the value of $(BD)^2$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile