90.2k views
5 votes
Urgent, basic level high school geo

Urgent, basic level high school geo-example-1

1 Answer

7 votes

The expression for ∠3 is x² + 2x, m∠3 is 48⁰ and ∠ SXP is 138⁰

How to find the value of x and ∠SXP

from the figure.

Given that QX is perpendicular to XS, then,

∠1 + ∠2 = 90⁰(complementary angles)

(x² + 2x) + (5x + 12) = 90⁰

x² + 7x + 12 = 9⁰

Subtract 90 from both sides to set the equation to zero:

x² + 7x + 12 - 90 = 0

Combine like terms: x² + 7x - 78 = 0

Factor the quadratic expression

: (x + 13)(x - 6) = 0

x + 13 = 0 orx - 6 = 0

x = -13 or x = 6

So, the solutions to the quadratic equation are (x = -13) and (x = 6).

Also,

PX is perpendicular to XR, then,

∠2 + ∠3 = 90⁰(complementary angles)

5x + 12 + ∠3 = 90⁰

Therefore

∠2 + ∠3 = ∠1 + ∠2

5x + 12 + ∠3 = (x² + 2x) + (5x + 12)

Solve for ∠3

∠3 = x² + 2x

Substitute x = 6 into the expressions

∠1 = x² + 2x

= 6² + 2(6)

= 36 + 12

= 48⁰

∠2 = 5x + 12

= 5(6) + 12

= 30 +12

= 42⁰

Then ∠3 is x² + 2x

∠SXP =∠1 + ∠2 + ∠3

= 48 + 42 +48

= 138⁰

User Isolated Ostrich
by
7.7k points