A gable roof (isosceles triangle-shaped) has a vertical height of 2.1 metres and the ceiling is 10.9 meters from one side to the other. Find the pitch (angle) of the roof.

Step-by-step explanation:

The given values can be mapped to an isosceles .

Side AB = AC

Vertical height, AD = 2.1 m

The distance between one side to the other side of ceiling = 10.9 m

To find:

Pitch (Angle of the roof ) = ?

i.e. or = ? (because it is isosceles triangle, so both will be equal)

Solution:

As is isosceles, so vertical height will divide the side BC in two equal parts

i.e.

In , let us use tangent trigonometric property.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

the first one

step-by-step explanation:

just id the test.