Question

fast
class 10
chapter 8or 9
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Answer 1

given,

See the attachment for diagram,

CD=10=BE

AB=height of tower=h

AE=AB-BE=h-10

let,BC=DE=x

solution,

In triangle ABC,)

tan60°=AB/BC=h/x=√3 (tan60=√3).........(i)

In triangle AED,

tan45°=AE/DE=(h-10)/x=1 (tan45=1)........(ii)

by (i) and (ii),

we got,

h=√3x and h=x+10

=>x+10=√3x

√3x-x=10

x(√3-1)=10

x=10/(√3-1)

x=10(√3+1)/(√3-1)(√3+1)

 =10(√3+1)/2

x=5(√3+1)

now,

x=h-10

5√3+5+10=h

5√3+15=h or

5√3(√3+1)=h

Answer 2

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SOLUTION :


=> Given ; I) the first angle is 45°


=> the hypotenuse is given 10 m


by using trigonometric values


=> we will take sin@ value


=> Sin@ = p/h


=> let's come to question


=> sin@45°=p/h


=> here we know P = x , and h = 10m


=> sin@45° = x/10


=>we also know that sin@45° = 1/√2


=> 1/√2 = x/10


=> √2x = 10


=> x = 10/√2 .


=> we write it as >>>>>


=> 2 × 5/√2


=> √2 × 5


=> 5√2 m


=> so, the height of that tower is equal to 5√2 metres.