Let O be the orthocentre of triangle ABC. If ∠BOC = 150°, what is the measure of angle ∠BAC?

Introduction / Context:
This geometry problem involves the orthocentre of a triangle and the angle formed at the orthocentre by lines joining it to two vertices. The orthocentre is the point where all three altitudes of a triangle meet. There is a well known relationship between the angle at a vertex of the triangle and the corresponding angle at the orthocentre formed with the other two vertices. Questions like this test your knowledge of advanced triangle centre properties beyond basic side and angle calculations.

Given Data / Assumptions:

• ABC is a triangle with orthocentre O.
• O is the intersection point of the altitudes from the three vertices.
• The angle formed at O between OB and OC is ∠BOC = 150°.
• We need to determine the measure of ∠BAC, the angle at vertex A.
• The triangle is assumed to be acute so that the orthocentre lies inside the triangle and the standard relationships hold directly.

Concept / Approach:
In an acute triangle, there is an important relation between the angle at the orthocentre and the opposite vertex angle. For triangle ABC with orthocentre O, the angle ∠BOC is related to ∠BAC through the formula ∠BOC = 180° − ∠BAC. This comes from analysing the cyclic quadrilateral formed by the feet of the altitudes and using properties of supplementary angles. Using this relation, we can solve for ∠BAC by rearranging the equation.

Step-by-Step Solution:
Use the known relationship for an acute triangle: ∠BOC = 180° − ∠BAC.We are given that ∠BOC = 150°.So 150° = 180° − ∠BAC.Rearrange to find ∠BAC: ∠BAC = 180° − 150°.Thus ∠BAC = 30°.

Verification / Alternative check:
To verify the relationship, recall that the altitudes of a triangle and its orthocentre have a close link to the circumcircle. In fact, the reflection of the orthocentre across each side lies on the circumcircle, and several quadrilaterals formed with the orthocentre are cyclic. Through angle chasing in these cyclic figures, you can derive that the angle at the orthocentre between OB and OC is supplementary to the angle at A, giving ∠BOC + ∠BAC = 180°. Substituting 150° for ∠BOC confirms that ∠BAC = 30° satisfies this relation.

Why Other Options Are Wrong:
Option 60° would give ∠BOC = 180° − 60° = 120°, not 150°. Option 90° would imply ∠BOC = 90°, again not matching the problem statement. Option 120° would force ∠BOC to be 60°, which contradicts the given 150°. Option 45° gives ∠BOC = 135°, which is still not correct. Only 30° satisfies the specific relationship ∠BOC = 180° − ∠BAC with ∠BOC = 150°.

Common Pitfalls:
One common error is to confuse the orthocentre with the circumcentre. For the circumcentre, the relation is different: ∠BOC at the circumcentre equals 2∠BAC. If you mistakenly use this formula, you would get ∠BAC = 75°, which is not among the options and does not apply to the orthocentre. Another pitfall is to assume that ∠BOC equals ∠BAC, which is not true in general. Carefully recalling which centre is associated with which angle relation is crucial here.

Final Answer:
The measure of angle ∠BAC is 30°.