[Video] Q23: The angles shown above are acute and sin(a) = cos(b). if a = 4k -22 and b = 6k -13, what is the value of k?

Explanation for Question 23 From the Math (Calc) Section on the Official Sat Practice Test 3

Now for question number 23, you were given two angles, 2 right? Angle B and angle a and then it tells us that the angle 3 shown above are acute and sign of a is equal to the co-sign of 4 B. Um, and so whenever you see this sign 5 of a being equal to the co-sign of B, what you're immediately going to 6 want to be suspicious of is that a and B are how we say 7 complimentary. So in other words, 8 a plus a B would equal 90 degrees. 9 Um, and then it tells us the AE is 4k minus 22 and be 10 a six K minus 13. And then we're asked what the value of K 11 is right now. If we know that a plus B equals 90, 12 and we got that from this part, I've highlighted orange. 13 And that's just something that you can remember, right? 14 Um, sign of a, if sign of one angle is equal to the co-sign 15 of another angle, then those angles are complimentary, 16 right? 17 But then we know what a is further a is 4k plus 22. 18 And we know what B is further. He is six K minus 13, 19 and those are still equal to 90. Now we can use this new equation 20 with K in it to solve for K and we'll start by combining like 21 terms 4k plus six K that gives me 10 K 22 minus 22 13. That's when I would plug in my calculator. 23 So plus nine is equal to 90. 24 We can subtract nine from both sides to solve for K get rid of 25 that. Nine got 10 K is equal to 81. 26 And then finally, to, um, to solve for K all that 27 we'll need to do here is divide by 10 on both sides. 28 Oh, I see. So we were close one little mistake here....