Question : In $\Delta ABC$, the external bisector of the angles, $\angle B$ and $\angle C$ meet at the point $O$. If $\angle A = 70^\circ$, then the measure of $\angle BOC$:
Option 1: $55^\circ$
Option 2: $75^\circ$
Option 3: $60^\circ$
Option 4: $50^\circ$
New: SSC CHSL tier 1 answer key 2024 out | 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: $55^\circ$
Solution : Given: $\angle A = 70^\circ$ To find: $\angle BOC$ We know that, Angle formed by external angle bisectors = $90^\circ-\frac{1}{2}\angle A$ ⇒ $\angle BOC$ = $90^\circ-\frac{1}{2}\angle A$ = $90^\circ-\frac{1}{2}×70^\circ$ = $90^\circ-35^\circ$ = $55^\circ$ Hence, the correct answer is $55^\circ$.
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 $\triangle ABC$, the internal bisectors of $\angle B$ and $\angle C$ meet at point $O$. If $\angle A = 80^\circ$, then $\angle BOC$ is equal to:
Question : $\triangle ABC$ is an isosceles triangle with AB = AC. If $\angle BAC=50^\circ$, then the degree measure of $\angle ABC$ is equal to:
Question : If the angles $P, Q$ and $R$ of $\triangle PQR$ satisfy the relation $2 \angle R-\angle P=\angle Q-\angle R$, then find the measure of $\angle R$.
Question : The internal bisectors of the angles B and C of a triangle ABC meet at I. If $\angle$BIC = $\frac{\angle A}{2}$ + X, then X is equal to:
Question : $O$ is the orthocentre of triangle $ABC$, and if $\angle BOC = 110^\circ$, then $\angle BAC$ will be:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile