How do you find the angles of a triangle whose side lengths are 50, 70, and 100?

Sep 7, 2016

111.8°, 40.5°, 27.7°

Explanation:

You are given all 3 sides of a non-right-angled triangle.
With all three sides we can us the Cos Rule.

Let's call the triangle $\Delta$PQR, with sides as $p = 100 , q = 50 \mathmr{and} r = 70$

It's a good idea to find the biggest angle first using the cos rule, because if it is obtuse, the cos value will indicate this, but the sin value will not.
The BIGGEST angle is OPPOSITE the LONGEST side.

$\cos P = \frac{{q}^{2} + {r}^{2} - {p}^{2}}{2 q r}$

cos P = (50^2 +70^2 - 100^2)/(2(50)(70)

$\cos P = - 0.3714 \text{larr" the minus sign means an obtuse angle}$

hatP = 111.8°" find arc cos"

Now use the sine rule, because you have a matched pair.
(Side p and angle P)

$\frac{\sin R}{r} = \frac{\sin P}{p}$

$\frac{\sin R}{70} = \frac{\sin 111.8}{100}$

$\sin R = \frac{70 \times \sin 111.8}{100} = 0.6499$

R= 40.5°" " find arcsin

Now from the sum of the angles in a triangle we can get

hatQ = 180° -111.8°-40.5° = 27.7°