Triangle Angle Bisector Problem
In triangle ABC, the bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ meet at O. Given $\angle\text{BAC} = 60^\circ$, find $\angle\text{BOC}$.
Triangle Angle Sum
The sum of angles in triangle ABC is $180^\circ$:
$ \angle\text{BAC} + \angle\text{ABC} + \angle\text{ACB} = 180^\circ $
Substitute $\angle\text{BAC} = 60^\circ$:
$ 60^\circ + \angle\text{ABC} + \angle\text{ACB} = 180^\circ $
Thus, the sum of the other two angles is $\angle\text{ABC} + \angle\text{ACB} = 180^\circ - 60^\circ = 120^\circ$.
Angle BOC Calculation
Consider triangle BOC. The sum of its angles is $180^\circ$:
$ \angle\text{OBC} + \angle\text{OCB} + \angle\text{BOC} = 180^\circ $
Since BO and CO are angle bisectors:
$ \angle\text{OBC} = \frac{1}{2} \angle\text{ABC} $
$ \angle\text{OCB} = \frac{1}{2} \angle\text{ACB} $
Substitute these into the triangle BOC angle sum equation:
$ \frac{1}{2} \angle\text{ABC} + \frac{1}{2} \angle\text{ACB} + \angle\text{BOC} = 180^\circ $
Factor out $\frac{1}{2}$:
$ \frac{1}{2} (\angle\text{ABC} + \angle\text{ACB}) + \angle\text{BOC} = 180^\circ $
Use the sum calculated earlier ($\angle\text{ABC} + \angle\text{ACB} = 120^\circ$):
$ \frac{1}{2} (120^\circ) + \angle\text{BOC} = 180^\circ $
$ 60^\circ + \angle\text{BOC} = 180^\circ $
Solve for $\angle\text{BOC}$:
$ \angle\text{BOC} = 180^\circ - 60^\circ = 120^\circ $
The measure of $\angle\text{BOC}$ is $120^\circ$. This corresponds to Option A.
Similar Questions
- In $\triangle ABC$, $BD \perp AC$ at $D$ and $\angle DBC = 40^\circ$. $E$ is a point on $BC$ such that $\angle CAE = 37^\circ$. What is the measure of $\angle AEB$?
- In $\Delta ABC$, $BD \perp AC$ at D and $\angle DBC = 71^\circ$. E is a point on BC such that $\angle CAE = 17^\circ$. What is the measure of $\angle AEB$?
- In $\Delta\text{XYZ}$, $\text{XY} = 12\text{ cm}$ and $\text{YZ} = 18\text{ cm}$. $\text{XW}$, the angle bisector of $\text{YXZ}$, meets $\text{YZ}$ at $\text{W}$, such that $\text{YW} : \text{WZ}$ is $4 : 5$. Find the length of the third side of the triangle.
- In $\Delta\text{PQR}$, $\text{QR}$ is extended up to $\text{S}$ so that $\text{RS} = \text{RP}$. If $\angle\text{PRQ} = 70^\circ$ and $\angle\text{QPS} = 110^\circ$ then find the measure of $\angle\text{PQS}$.
- The areas of two similar triangles are respectively $16\text{ m}^{2}$ and $36\text{ m}^{2}$. Find the ratio of their corresponding sides.
- In $\Delta PQR$, QR is extended up to S, so that RS = RP. If $\angle RPQ = 55^\circ$ and $\angle PRS = 110^\circ$, then find the measure of $\angle PQS$.
- The sides of a triangle are 44 cm, 33 cm, and 55 cm. What is its area? (in $cm^2$)
- The perimeter of an equilateral triangle ABC is 22.2 cm. What is the area of triangle (in $cm^2$)?
- In $\triangle\text{ABC}$, $\text{BD} \perp \text{AC}$ at $\text{D}$ and $\angle\text{DBC} = 60^{\circ}$. $\text{E}$ is a point on $\text{BC}$ such that $\angle\text{CAE} = 20^{\circ}$. What is the measure of $\angle\text{AEB}$?
- In a triangle, if angle $A = 30^\circ$ and angle $B = 45^\circ$, then what is the angle of C?
Important Questions from Triangles
What is the circumcenter of the triangle ABC?
What is the centroid of the triangle ABC?
What is the foot of the altitude from the vertex A of the triangle ABC?
In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?
In ΔABC, ∠A = 66° and ∠B = 50 °. If the bisectors of ∠B and ∠C meet at P, then ∠BPC – ∠PCA = ?
