Question : In a triangle $ABC$, side $BC$ is produced to $D$ such that $\angle A C D=127^{\circ}$. If $\angle A B C=35^{\circ}$, then find $\angle B A C$.
Option 1: 82°
Option 2: 92°
Option 3: 95°
Option 4: 75°
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: 92°
Solution : Given, In a triangle $ABC$, $\angle A C D=127^{\circ}$ and $\angle A B C=35^{\circ}$ By exterior angle property, $\angle A C D = \angle A B C+\angle B A C$ $⇒127^{\circ} = 35^{\circ}+\angle B A C$ $⇒\angle B A C=127^{\circ} - 35^{\circ}$ $\therefore\angle B A C=92^{\circ}$ Hence, the correct answer is 92º.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : The side BC of a triangle ABC is produced to D. If $\angle ACD = 112^\circ$ and $\angle B =\frac{3}{4} \angle A$ then the measure of $\angle B$ 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 : $\triangle ABC$ is an equilateral triangle. The side $BC$ is produced to point $D$. If $A$ joines $D$ and $B C=CD$, then the degree measure of angle $CAD$ is equal to:
Question : In a $\triangle$ ABC, BC is extended to D and $\angle$ ACD = $120^{\circ}$. $\angle$ B = $\frac{1}{2}\angle$ A. Then $\angle$ A is:
Question : The side $BC$ of a triangle $ABC$ is extended to $D$. If $\angle ACD = 120^{\circ}$ and $\angle ABC = \frac{1}{2} \angle CAB$, then the value of $\angle ABC$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile