Explanation:
So in this problem, we're looking at a right-angled triangle where angle C is 90 degrees. We need to find the remaining sides and angles of this triangle given that side b is 20 units long and side c is 40 units long. To do this, we can use some basic trigonometry. Let's start with finding angle A. Using the sine ratio, we can say that sin(A) = b/c. Plugging in the values, we have sin(A) = 20/40 which simplifies to 1/2. Taking the inverse sine of both sides, A = arcsin(1/2). Now if you remember your special angles, this gives us A = 30 degrees. To find side a, we can use Pythagoras' theorem: a^2 + b^2 = c^2. Plugging in the values gives us a^2 + 20^2 = 40^2 which simplifies to a^2 + 400 = 1600. Solving for a gives us a ≈ √1200 ≈ 34 (rounded to the nearest ten). So there you have it - angle A is approximately 30 degrees, side a is approximately 34 units long, and we have successfully solved for angle A and side a in this right-angled triangle problem. Now, let's move on to finding angle B. Since angle C is 90 degrees, the sum of angles A and B must equal 90 degrees as well. Therefore, angle B = 90 - A. Substituting the value of A as 30 degrees, we find that angle B = 90 - 30 = 60 degrees. So, angle B is approximately 60 degrees. Now that we have determined all the angles and sides of this triangle, we can confidently say that in this right-angled triangle with side b measuring 20 units and side c measuring 40 units, angle A is approximately 30 degrees, angle B is approximately 60 degrees, side a is approximately 34 units long, and angle C is 90 degrees.