Question : In triangle ABC, $\angle$ABC = 15°. D is a point on BC such that AD = BD. What is the measure of $\angle$ADC (in degrees)?
Option 1: 15
Option 2: 30
Option 3: 45
Option 4: 60
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 30
Solution : Given: In triangle ABC, $\angle$ABC = 15°. D is a point on BC such that AD = BD. Given that AD = BD. Then, $\angle$ABD = $\angle$BAD = 15° Now, in $\triangle$ABD ⇒ $\angle$ABD + $\angle$BAD + $\angle$ADB = 180° ⇒ $\angle$ADB = 180° – 15° – 15° ⇒ $\angle$ADB = 150° We know that, $\angle$ADB + $\angle$ADC = 180° ⇒ $\angle$ADC = 180° – 150° ⇒ $\angle$ADC = 30° Hence, the correct answer is 30.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : In $\triangle $ABC, AD$\perp$ BC and AD2 = BD × DC. The measure of $\angle$ BAC is:
Question : In $\triangle ABC$, $\angle B=60°$, $\angle C=40°$. AD is the bisector of $\angle A$ and AE is drawn perpendicular on BC from A. Then the measure of $\angle EAD$ is:
Question : $ABC$ is an isosceles triangle with $AB = AC$, The side $BA$ is produced to $D$ such that $AB = AD$. If $\angle ABC = 30^{\circ}$, then $\angle BCD$ is equal to:
Question : In $\triangle$ABC, $\angle$A = $\angle$B = 60°, AC = $\sqrt{13}$ cm, the lines AD and BD intersect at D with $\angle$D = 90°. If DB = 2 cm, then the length of AD is:
Question : $\angle A$ of $\triangle ABC$ is a right angle. $AD$ is perpendicular on $BC$. If $BC= 14$ cm and $BD= 5$ cm, then measure of $AD$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile