If the bottom of a ship is 15.8 below the surface of the sea, calculate the pressure at the bottom of the ship. The density of t
1 answer:
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Answer:
x= 37.5°
Step-by-step explanation:
∠CBD
= 180° -75° (adj. ∠s on a str. line)
= 105°
∠BCD= ∠BDC (base ∠s of isos. △BCD)
∠BCD= x
∠BCD +∠BDC +∠CBD= 180° (∠ sum of △BCD)
x +x +105°= 180°
2x= 180° -105°
2x= 75°
x= 37.5°
<u>Alternative</u><u> </u><u>working</u><u>:</u>
∠BDA= (180° -75°) ÷2 (base ∠s of isos. △ABD)
∠BDA= 52.5°
∠BDA +∠BDC= 90°
52.5° +x= 90°
x= 90° -52.5°
x= 37.5°
