Posted by **anon** on Monday, January 10, 2011 at 10:09pm.

In a right triangle sinA= 3/5 and C=17. Find a and b.

Be sure to draw a right triangle and letter it properly.

how can i solve the math in this problem. i know i would have to use the Pythagoream Theorem but i dont know how to set the problem up first.

- trigonometry. -
**bobpursley**, Monday, January 10, 2011 at 10:19pm
SinA= b/C=3/4

you are given C, so b= 3/4 * 17

then a can be found from a^2+b^2=C^2

- trigonometry. -
**Reiny**, Monday, January 10, 2011 at 10:22pm
your question makes no sense

If you have a right-angled triangle where

sinA = 3/5, then the opposite side to vertex A is 3 and the hypotenuse is 5,

by Pythagoras it is easy to show that the third side is 4

(you should have recognized the 3-4-5 right-angled triangle)

if sinA = 3/5, then angle A is appro 37.9°

what does C=17 mean?

- trigonometry. -
**anon**, Monday, January 10, 2011 at 11:52pm
sinA= 3/5 and C=17

Finding a and b

would this work as an answer ..?

sinA = opp/hyp = a/c = 3/5 = 3*3.4/5*3.4 = 10.2/17

a = 10.2

Applying Pythagoream Theorem a^2 + b^2 = c^2

- trigonometry. -
**anon**, Monday, January 10, 2011 at 11:55pm
b= sqrt 17^2 - 10.2^2 = 13.6

??

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