A 28-foot ladder is placed against a building and the top of the ladder makes a 27° angle with the building. How far away from the building is the base of the ladder? Write only the number rounded
to the nearest tenth of a foot.

Step-by-step explanation:

The set up is a right angles triangle

Given

The length of the ladder = 28 foot (hypoenuse)

theta = angle of elevation = 27 degrees

Required

Distance of the building from the base of the ladder (Adjacent)

According to CAH;

cos theta = adj/hyp

cos 27 = adj/28

adj = 28cos27

adj = 24.95 foot

Hence the length of the building from the base of the ladder is 24.95 feet

-81 -56 -27 0 81

step-by-step explanation:

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.