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 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 = ?
In a triangle ABC, points P and Q are on AB and AC, respectively, such that AP = 4 cm, PB = 6 cm, AQ = 5 cm and QC = 7.5 cm. If PQ = 6 cm, then find BC (in cm).
The perimeters of two similar ΔABC and Δ PQR are 48.4 cm and 12.1 cm, respectively. What is the ratio of the areas of Δ ABC and Δ PQR?
