# Thread: A couple triangle + angle value problems...

1. ## A couple triangle + angle value problems...

Hey everybody! I've been working on this huge assignment and I've got everything done except for these two questions which are completely stumping me for some odd reason... so yeah, help would be extremely appreciated. :]

1.) In △ABC, AC = 7, BC = 12, and ∠B = 35°. Solve the triangle?

2.)

Find the values of angle Θ such that 0° <= Θ <= 360° for the following trig functions?

a) sin Θ = 0.64217

b) cos Θ = -0.42613

c) tan Θ = 1.86721

Sorry #2 looks like it's in bold for some reason... I was playing around with the fonts and it went like that and I can't change it, haha. Anyways, yeah, thanks so much in advance everyone!

2. Hello, Shnub!

1) In $\Delta ABC\!:\;\;b = 7,\; a = 12,\;\angle B = 35^o$
. . . Solve the triangle.
Code:
                  A
*
*  *
*     *
c *        * b = 7
*           *
* 35°          *
B *  *  *  *  *  *  * C
a = 12

Law of Sines: . $\frac{\sin A}{a} \:=\:\frac{\sin B}{b} \quad\Rightarrow\quad \frac{\sin A}{12} \:=\:\frac{\sin35^o}{7}$

. . . . . . . . . . $\sin A \:=\:\frac{12\sin35^o}{7} \;=\;0.983273891$

. . . . . . . . . . $A \;=\;79.50597013^o \quad\Rightarrow\quad \boxed{A \;\approx\;70.5^o}$

Then: . $C \;=\;1870^o - 35^o - 79.5^o \quad\Rightarrow\quad \boxed{C \;=\;65.5^o}$

Law of Sines: . $\frac{c}{\sin C} \:=\:\frac{b}{\sin B} \quad\Rightarrow\quad \frac{c}{\sin65.5^o} \:=\:\frac{7}{\sin35^o}$

. . . . . . . . . . $c \:=\:11.10528343 \quad\Rightarrow\quad \boxed{c \;\approx\;11.1}$