Two of the exterior angles of a triangle are $158^\circ$ and $99^\circ.$ Find the third exterior angle, in degrees.

Mathematics

1 answer:

Lisa [10]3 years ago

4 0

Answer:

The third exterior angle is $103^o$

Step-by-step explanation:

Given

$\theta = 158^o$

$\alpha = 99^o$

Required

The third exterior angle $(\beta)$

The exterior angles of a triangle add up to 360 degrees.

So:

$\theta + \alpha + \beta = 360^o$

Make $\beta$ the subject

$\beta = 360^o - (\theta + \alpha)$

Substitute known values

$\beta = 360^o - (158^o + 99^o)$

$\beta = 103^o$

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Help find zeros for 9 and 10

Bingel [31]

x4-10x2+9=0 Four solutions were found : x = 3 x = -3 x = 1 x = -1

Step by step solution :Step 1 :Skip Ad
Equation at the end of step 1 : ((x4) - (2•5x2)) + 9 = 0 Step 2 :Trying to factor by splitting the middle term

2.1 Factoring x4-10x2+9

The first term is, x4 its coefficient is 1 .
The middle term is, -10x2 its coefficient is -10 .
The last term, "the constant", is +9 

Step-1 : Multiply the coefficient of the first term by the constant 1 • 9 = 9

Step-2 : Find two factors of 9 whose sum equals the coefficient of the middle term, which is -10 .

-9 + -1 = -10 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -9 and -1
 x4 - 9x2 - 1x2 - 9

Step-4 : Add up the first 2 terms, pulling out like factors :
 x2 • (x2-9)
 Add up the last 2 terms, pulling out common factors :
 1 • (x2-9)
Step-5 : Add up the four terms of step 4 :
 (x2-1) • (x2-9)
 Which is the desired factorization

Trying to factor as a Difference of Squares :

2.2 Factoring: x2-1

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =
 A2 - AB + BA - B2 =
 A2 - AB + AB - B2 =
 A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 1 is the square of 1
Check : x2 is the square of x1 

Factorization is : (x + 1) • (x - 1)

Trying to factor as a Difference of Squares :

2.3 Factoring: x2 - 9

Check : 9 is the square of 3
Check : x2 is the square of x1 

Factorization is : (x + 3) • (x - 3)

Equation at the end of step 2 : (x + 1) • (x - 1) • (x + 3) • (x - 3) = 0 Step 3 :Theory - Roots of a product :

3.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.