trigo help pls?

Question

given a right triangle ... (triangle ABC .... C= hypotenuse)

solve for A

1. sin 2A = cos 13A

Answer 1

ans. A=6
xplanation:
Since sin and cosine are cofunctions, their functions can only be equal if the sum of their respective angles is 90 degrees, in a right triangle (take note that the total measures of the angles of a triangle is 180 and for a right triangle the angle opposite the hypotenuse is the right angle with a measure of 90 degrees, so the sum of the two other angles is 90 degrees) . Hence, if 2A stands for the first acute angle, (90 -2A) stands for the other acute angle. Solving the problem we have:

sin 2A = cos 13A

but 13A = 90 -2A
therefore:

13A+2A=90
15A=90
A=90/15
A=6 degrees

Answer 2

You asked for 3 answers, here are 4:
6 deg
30 deg
54 deg
78 deg
Done by method similar to answer 3 but setting 13A + 2A equal to not just 90 deg but 1, 2, 3... full revolutions + 90 deg (with the limit A < 90 deg since it's acute by definition). Thus the 13A + 2A values that yield the desired sin/cos ratio are 90, 450, 810 and 1170 deg, i.e. (0, 1, 2 and 3) revolutions + 90 deg, so you can successively add 24 deg to the original 6 deg to get the additional solutions.

Answer 3

LET SIN=D AND COS=X
THERE FOR
SIN 2A=COS13A=D GENRATION X
HENCE PROVE.

Answer 4

what is sin/cos =? you do the rest