in a right sin A= 3/4 and c= 17 find a and b. be sure to draw a right triangle and letter it properly.?

Answer 1

i'm not sure if my answer is right but i think this is how u do it.

SIN A = 3/4
SIN = opposite/hypotenuse
opposite is a = 3
hypotenuse is c = 4....but im not sure, cuz u said that c = 17

Answer 2

In a standard rt angle triangle (RAT) angle C=90 deg and side c is the hypotenuse. Here c=17 and angle A is across from side a. sin A= 3/4 = a/c = a / 17
so a = 17*3/4= 12.75 is the base of the RAT is ca and is called b and is 11.2 units long. Angle C=90 deg side cb is the vertical side which is 12.75 units long. the slanted side is c and is opposite the right angle. Angles are in CAPS and sides are small letters.

...B
a |\ c
...|_\
....b A

Answer 3

let the right angle triangle be ABC, AB, vertical and BC horizontal, with them at right angle. AC is the hypothenese

then angle A is BAC a=BC b=AC c=AB
sinA = 3/4 = a/b
c=17 and
c^2 + a^2 = b^2
17^2 + a^2 = b^2
289 + a^2 = b^2 - - - - (1)
3b = 4a
9b^2 = 16a^2 - - - - - - (2)
289 + a^2 = (16/9)a^2
(7/9)a^2 = 289
a^2 = 289(9/7) = 371.57142857142856
a = 19.27618812347059
b = 4a/3 = 25.70158416462745

Answer 4

sinA=3/4=y/h, where y is the height, and h is the hypotenuse, or in your case, c.

So

sinA=3/4=y/17, so y=3*17/4=12.75

Then x^2=c^2-y^2=17^2-12.75^2=289-162.5625=126.4375

So x=sqrt(126.4375)=11.24444

Because (3/4)^2 is greater than 0.5, we know that sinA is larger than cosA. Thus, y we can label as b on the vertical side, and x we can label as a on the bottom.