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
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 don't know how to set the problem up first.
4 answers
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?
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?
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
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
b= sqrt 17^2 - 10.2^2 = 13.6
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2 answers