Question

If measure of angle p=(10x-2),measure of angle o=(3x+9) , and measure of angle q=(3x-3) , list the lengths of the sides of triangle OPQ from longest to shortest.

Answer 1

The sum of all angles of a triangle must be equal to 180°

So we can find the value of x...

10x-2+3x+9+3x-3=180

16x+4=180

16x=176

x=11

So the angles are 108°, 42°, 30°

Now using the Law of Sines we can solve for all sides.

(sina/A=sinb/B=sinc/C)

a/sin30=b/sin42=c/sin108

However, we would need to know at least one side length to solve for sides a,b, and c numerically, otherwise it can be any multiple of an arbitrary choice for the smallest side.  Let a=1 for example:

b=sin42/sin30, c=sin108/sin30

b≈1.34,  c≈1.90

a=1, b=1.34, c=1.9

Anyway...

If you are given any side length:

a/sin30=b/sin42=c/sin108 still holds true and you can solve for the other side lengths...

Answer 2

Answer:

OQ > PQ > OP

Step-by-step explanation:

In this question measure of all angles in a triangle has been given.

m∠ p = (10x - 2)

m∠ q = (3x - 3)

m∠ o = (3x + 9)

Since total of all angles is 180°, so we will equate the total of all angles to 180°

∠p + ∠q + ∠o = 180°

(10x - 2) + (3x - 3) + (3x + 9) = 180°

(10x + 3x + 3x) -2 - 3 + 9 = 180

16x - (2 + 3 - 9) = 180°

16x + 4 = 180

16x = 180 - 4

16x = 176

x = $\frac{176}{16}=11$

Now we will find the measure of each angle.

∠p = (10x - 2) = 10×11 - 2 = (110 - 2) = 108°

∠q = (3x - 3) = 3×11 - 3 = (33 - 3) = 30°

∠o = (3x + 9) = 3×11 + 9 = (33 + 9) = 42°

As we know in a triangle, angle opposite to the largest side is largest and angle opposite side is the smallest.

Since ∠p > ∠o > ∠q

Therefore, in Δ OPQ opposite sides to these angles will be in the same ratio.

OQ > PQ > OP