SOLUTION: if sin A=6÷13 and SecB=13÷6 then A+B is equal to 150° 90° 60° 45°

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Question 1133873: if sin A=6÷13 and SecB=13÷6 then A+B is equal to
150°
90°
60°
45°
Found 3 solutions by MathLover1, MathTherapy, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

if sin+%28A%29=6%2F13=>A=sin+%5E-1%286%2F13%29=>A=27.49°
and if sec%28B%29=13%2F6+=>sec%28B%29=1%2Fcos%28B%29+
=>1%2Fcos%28B%29=13%2F6+
=>cos%28B%29=1%2F%2813%2F6%29
=>cos%28B%29=6%2F13
B=cos%5E-1%286%2F13%29
B=62.51°

then A%2BB=27.49%2B62.51°
A%2BB=90°


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
if sin A=6÷13 and SecB=13÷6 then A+B is equal to
150°
90°
60°
45°




As seen, x = y = 6, and so, this is an ISOSCELES right triangle. 

This means that ∡s A and B are 45o each, and 2(45o) = highlight_green%2890%5Eo%29

NO NEED to CALCULATE ANYTHING!!

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There is indeed no need to calculate anything; but the two angles are not each 45 degrees. (sin(45) is not 6/13....)

We are given sinA = 6/13 and secB = 13/6. We then know that cosB = 6/13.

So we have an angle A whose sine is the same as the cosine of another angle B.

Picture a right triangle with A and B as the acute angles. Then the basic definition of sine and cosine tells us that the sine of one of the angles is the cosine of the other.

So, because we have two angles with the sine of one being equal to the cosine of the other, we know the angles are the acute angles of a right triangle; therefore their sum is 90 degrees.