Find the value of given expression
2 answers:
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Answer:
Given A triangle ABC in which
∠C =90°,∠A=20° and CD ⊥ AB.
In Δ ABC
⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]
⇒20° + ∠B + 90°=180°
⇒∠B+110° =180°
∠B =180° -110°
∠B = 70°
In Δ B DC
∠BDC =90°,∠B =70°,∠BC D=?
∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]
90° + 70°+∠BC D =180°
∠BC D=180°- 160°
∠BC D = 20°
In Δ AC D
∠A=20°, ∠ADC=90°,∠AC D=?
∠A + ∠ADC +∠AC D=180° [angle sum property of triangle]
20°+90°+∠AC D=180°
110° +∠AC D=180°
∠AC D=180°-110°
∠AC D=70°
So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°
Answer:
Step-by-step explanation:
let the speed of one plane=x
speed of other plane=x+80
time=4 hrs
then 4x+4(x+80)=3800
4x+4x+320=3800
8x=3800-320
or 8x=3480
x=435
speed of one plane=435 m/h
speed of other plane=435+80=515 m/h
2.
Let he walks x hrs
then he take a taxi for 2-x hrs
4x+(2-x)50=31
4x+100-50x=31
-46x=-69
so he walks 3/2 hrs=1.5 hrs
travels by taxi=2-1.5=0.5 hrs
distance traveled by taxi=0.5×50=25 miles