# How do you solve the right angle triangle given A=70°-49'-15.6", a=23, b=8?

##### 1 Answer
Jan 14, 2017

angleB=19°10'44.4'', c = 24.352

#### Explanation:

An easy way to remember the trigonometric ratios in right angled
triangles:

Let opposite side be “t”
Let hypotenuse side be “s”
Let adjacent side be “a”

$S \in \theta = \frac{o p p o s i t e}{h y p o t e \nu s e} = \frac{t}{s} = S \in \left(t s\right)$

$C o s \theta = \frac{a \mathrm{dj} a c e n t}{h y p o t e \nu s e} = \frac{a}{s} = C o s \left(a s\right)$

$T a n \theta = \frac{o p p o s i t e}{a \mathrm{dj} a c e n t} = \frac{t}{a} = T a n \left(t a\right)$

$C o \sec \theta = \frac{h y p o t e \nu s e}{a \mathrm{dj} a c e n t} = \frac{s}{t} = C o \sec \left(s t\right)$

$S e c \theta = \frac{h y p o t e \nu s e}{a \mathrm{dj} a c e n t} = \frac{s}{a} = S e c \left(s a\right)$

$C o t \theta = \frac{a \mathrm{dj} a c e n t}{o p p o s i t e} = \frac{a}{t} = C o t \left(a t\right)$

180° -(70° 49'15.6''+90°)=19°10'44.4''= angleB

s/a = Sec 19°10'44.4''

Multiply both sides by a

s = Sec 19°10'44.4'' xx a

$s = 1.058764789 \times 23$

$s = 24.352$ = hypotenuse = side b

Check:

t/s = Sin 19°10'44.4'' 

Multiply both sides by s

t= Sin 19°10'44.4'' xx s

$t = 0.328520492 \times 24.352$

$t = 8$ = opposite side = b