2.1 If \( 12 \tan B-5=0 \) and \( 90^{\circ} \leq B \leq 360^{\circ} \), determine the value of \( \sin B+\cos B \) with the aid of a diagram.

We start with the given equation:   12 tan B – 5 = 0  ⟹  tan B = 5/12. Since tan B is positive and B is restricted to 90° ≤ B ≤ 360°, we must decide in which quadrant B lies. Recall that the tangent function is positive in the 1st and 3rd quadrants. However, the 1st quadrant (0° to 90°) is excluded by the lower bound (B must be at least 90°), so B must lie in the 3rd quadrant. In the 3rd quadrant, both sine and cosine are negative. Step 1. Draw a right triangle for the reference angle α, where   tan α = 5/12. Using the Pythagorean theorem, the hypotenuse is:   r = √(5² + 12²) = √(25 + 144) = √169 = 13. So, in the reference triangle:   Opposite side = 5,   Adjacent side = 12,   Hypotenuse = 13. Step 2. Determine sine and cosine of the reference angle:   sin α = 5/13, cos α = 12/13. Since B is in the 3rd quadrant, the actual values for B are:   sin B = –sin α = –5/13,   cos B = –cos α = –12/13. Step 3. Calculate sin B + cos B:   sin B + cos B = (–5/13) + (–12/13) = –(5 + 12)/13 = –17/13. Thus, the required value of sin B + cos B is –17/13. Diagram (Sketch Guidelines):  1. Draw the coordinate axes.  2. Mark a point in the 3rd quadrant corresponding to the terminal side of angle B = 180° + α (with α = arctan(5/12)).  3. Construct a right triangle by dropping a perpendicular from that point to the x-axis.  4. Label the sides using the reference triangle:   • The side opposite the angle (inside the triangle) with length 5,   • The side adjacent with length 12,   • Hypotenuse with length 13.  5. Note that since the triangle is in the 3rd quadrant, both the horizontal (x) and vertical (y) sides are negative. This completes the solution and supports our result. Final Answer: sin B + cos B = –17/13.