2.4.4 Journal: measurement and units answers because it’s a waist of time part 2
2 answers:
You might be interested in
Answer:
The distance of the overpass above the ground is approximately 26.795 ft
Step-by-step explanation:
The parameters given are;
The distance from the overpass the engineer stands before determining the angle of elevation of the overpass from his standing point = 100 ft
The angle of elevation of the overpass as determined by the engineer from 100 ft = 15°
By trigonometric ratios, we have;
The opposite side to the 15° angle of elevation in the above case is the distance of the overpass above the ground
The opposite side to the 15° is the distance of the engineer from the base of the overpass
Therefore;
Tan(15°) the height of the overpass=
length
The distance of the overpass above the ground = 100 × tan (15°) ≈ 26.795 ft.
<u>Solution-</u>
Given that,
In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.
Then considering ΔPQT and ΔSTF,
1- ∠FTS ≅ ∠PTQ ( ∵ These two are vertical angles)
2- ∠TFS ≅ ∠TPQ ( ∵ These two are alternate interior angles)
3- ∠TSF ≅ ∠TQP ( ∵ These two are also alternate interior angles)
<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>
∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.
As QS = TQ + TS = 10 (given)
If TS is x, then TQ will be 10-x. Then putting these values in the equation
∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm
