answer:1. **Given:** A triangle (ABC) with the sides and angle: - (AC = sqrt{2}) - (BC = 1) - (angle B = 45^circ) 2. **Apply the Law of Sines:** According to the Law of Sines, we have: [ frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} ] Here, angle (A) and side (a) (opposite angle (A)), angle (B) and side (b) (opposite angle (B)), and angle (C) and side (c) (opposite angle (C)). For our triangle: [ frac{BC}{sin angle A} = frac{AC}{sin angle B} ] Substituting the known values, we get: [ frac{1}{sin angle A} = frac{sqrt{2}}{sin 45^circ} ] 3. **Evaluate (sin 45^circ):** We know that: [ sin 45^circ = frac{sqrt{2}}{2} ] Substitute this value into the equation: [ frac{1}{sin angle A} = frac{sqrt{2}}{frac{sqrt{2}}{2}} = 2 ] 4. **Solve for (sin angle A):** [ sin angle A = frac{1}{2} ] 5. **Determine (angle A):** The solutions to the equation (sin theta = frac{1}{2}) are: [ theta = 30^circ quad text{or} quad theta = 150^circ ] 6. **Consider the sum of angles in a triangle:** We know that the sum of the angles in any triangle is (180^circ). Given (angle B = 45^circ), let's check the possible values for (angle A). - If (angle A = 30^circ): [ angle C = 180^circ - angle A - angle B = 180^circ - 30^circ - 45^circ = 105^circ ] - If (angle A = 150^circ): [ angle C = 180^circ - angle A - angle B = 180^circ - 150^circ - 45^circ = -15^circ quad (text{which is impossible}) ] 7. **Conclusion:** The only valid solution that satisfies the conditions of the problem is: [ angle A = 30^circ ] # Conclusion: [ boxed{30^circ} ]

answer:1. **Calculate the gas usage for the initial 400 miles**: [ frac{400 text{ miles}}{25 text{ miles per gallon}} = 16 text{ gallons} ] Karl used all his initial fuel in the first part of his journey. 2. **After the purchase of 10 gallons of gasoline**, Karl refuels to 10 gallons, since his tank was empty. 3. **Calculate how much gas is left upon arrival**, knowing that the tank contains one-third of its 16-gallon capacity: [ frac{1}{3} times 16 text{ gallons} = frac{16}{3} text{ gallons} ] Since Karl refilled 10 gallons, the gas used after refueling to the moment of arrival is: [ 10 text{ gallons} - frac{16}{3} text{ gallons} = frac{14}{3} text{ gallons} ] 4. **Convert this gas usage to miles**: [ frac{14}{3} text{ gallons} times 25 text{ miles per gallon} = frac{350}{3} text{ miles} approx 116.67 text{ miles} ] 5. **Calculate the total distance** traveled by adding the two segments: [ 400 text{ miles} + frac{350}{3} text{ miles} = frac{1550}{3} text{ miles} approx 516.67 text{ miles} ] Karl drove a total distance of 517 miles that day. (Rounding to nearest mile) The final answer is boxed{B) 517 miles}

answer:1. **Identify the Altitudes and Points**: - Let ( H_1, H_2, H_3 ) be the orthocenters of ( triangle ABC, triangle AKL, ) and ( triangle AMN ) respectively. 2. **Important Condition**: - We are given ( frac{BK}{KM} = frac{CL}{LN} ). 3. **Construct Perpendicular Lines**: - Draw ( AC )'s perpendicular line through ( M ) and let it intersect line ( H_1H_2 ) at ( H_3' ). - Draw ( AB )'s perpendicular line through ( N ) and let it intersect line ( H_1H_2 ) at ( H_3'' ). 4. **Identify Parallel Vectors**: - Since ( overrightarrow{H_3} perp overrightarrow{AC} ), ( overrightarrow{KH_2} perp overrightarrow{AC} ), and ( overrightarrow{BH_1} perp overrightarrow{AC} ), we have: [ overrightarrow{MH_3'} parallel overrightarrow{KH_2} parallel overrightarrow{BH_1} ] 5. **Calculate Vector Relations**: - Let’s express the vector relationship. Since: [ overrightarrow{H_2H_3'} = frac{KM}{BK} overrightarrow{H_1H_2} ] - Similarly, from the perpendicular ( AB ): [ overrightarrow{H_2H_3''} = frac{LN}{CL} overrightarrow{H_1H_2} ] 6. **Utilize Given Condition**: - Given ( frac{BK}{KM} = frac{CL}{LN} ), we can equate the proportions: [ frac{KM}{BK} = frac{LN}{CL} ] - Hence, [ overrightarrow{H_2H_3'} = overrightarrow{H_2H_3''} ] 7. **Intersection Point**: - This means ( H_3' ) and ( H_3'' ) are the same point because their vectors are equal: [ H_3' = H_3'' ] 8. **Conclusion**: - Therefore, the orthocenters ( H_1, H_2, ) and ( H_3 ) are collinear. [ boxed{} ]

answer:To find the number of months for which the interest is calculated, we can use the simple interest formula: Simple Interest (SI) = Principal (P) * Rate (R) * Time (T) Where: - SI is the simple interest - P is the principal amount - R is the annual interest rate (in decimal form) - T is the time in years Given: - SI = 400 - P = 10000 - R = 4% per annum = 0.04 (as a decimal) We need to find T in years, but since we want to find the number of months, we will convert T to months later. First, let's rearrange the formula to solve for T: T = SI / (P * R) Now, plug in the given values: T = 400 / (10000 * 0.04) T = 400 / 400 T = 1 year Since we want to find the number of months, we convert 1 year into months: 1 year = 12 months Therefore, the interest is calculated for boxed{12} months.