Question

alpha and beta are quadrant 1 angles with tan (alpha) = 8/15 and sec (beta) = 41/40. find sin (alpha + beta)

Answer 1

Pythagorean theorem for any right triangle... and
sin (alpha + beta) = sin alpha cos beta + sin beta cos alpha

Answer 2

for alpha, your tan is 8/15, meaning opposite side o=8 and adjacent side a=15, therefore \[h=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\]

Answer 3

you can now solve sin alpha and cos alpha based on function definitions...
sin alpha = opposite/hypotenuse
cos alpha = adjacent/hypotenuse

Answer 4

same procedure for the beta....
sec beta = hypotenuse/adjacent = 41/40....
which means opposite is missing, still you can solve it by Pythagorean Theorem...
after solving opposite... you can solve for sin beta
sin beta = opposite/hypotenuse
for cos beta = 1/sec beta = 40/41

Answer 5

once you get all of sin & cos of both alpha & beta...

Answer 6

so I add 17 and 57.3

Answer 7

no we'll use the identity
\[\sin (\alpha+\beta)=\sin \alpha \cos \beta + \sin \beta \cos \alpha\]

Answer 8

\[\sin \alpha = \frac{ 8 }{ 17 }\]

Answer 9

ok...got that

Answer 10

\[\cos \alpha = \frac{ 15 }{ 17 }\]

Answer 11

yes

Answer 12

\[\cos \beta = \frac{ 40 }{ 41 }\]

Answer 13

41/40?

Answer 14

for opposite side to angle beta.... we know h=41 and a=40 so
\[o =\sqrt{41^2-40^2}=\sqrt{1681-1600}=\sqrt{81}=9\]

Answer 15

yes

Answer 16

since sec beta = 1/cos beta...

Answer 17

\[\sin \beta =\frac{ 9 }{ 41 }\]

Answer 18

\[\sin(\alpha+\beta)=\frac{ 8 }{ 17 }\frac{ 40 }{ 41 }+\frac{ 9 }{ 41 }\frac{ 15 }{ 17 }\]

Answer 19

you can proceed for simplification... :)

Answer 20

do I multiply the first two numbers and then add them?

Answer 21

Multiply first then add...

Answer 22

great! you're the best.

Answer 23

not really, just a Math enthusiast... :)