Q-36. From the top of tower 100 m high, a man observes two cars on the
opposite sides of the towers ad in same straight lines with its base, with
angles of depression 30° and 45°. Find the distance between the cars.(
Take V3 = 1.732)
Q-37. State and prove Pythagoras theorem.
Q-38. If the pth term of an A. P. is q and q th term is p. Prove that nth term
is (p+q-n)
Q-39. If sin? @ = 3 Sin Cos .Then Prove that tan 0 = 1 or 2
Q-40. How Many numbers lie between 10 and 300 which when divided by 4
leave.
Answers
Step-by-step explanation:
It's the answer. Make me as brainest
Answer 36,
Let the Tower be AC = 100m
Distance between cars is BD
BD=BC+CD
Let the BC be 'x' and CD be 'y'
In Traingle ACD
Cot45°= Base/Perpendicular
Cot45°=y/100
1=y/100
y=100m
In Traingle ACB
Cot30°=x/100
1.732=x/100
1.732×100=x
x=173.2m
Distance between cars = BC+CD
=x+y
=173.2+100
=273.2m
Answer 37,
Pythagoras' theorem :-
→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Step-by-step explanation:
It's prove :-
➡ Given :-
→ A △ABC in which ∠ABC = 90° .
➡To prove :-
→ AC² = AB² + BC² .
➡ Construction :-
→ Draw BD ⊥ AC .
➡ Proof :-
In △ADB and △ABC , we have
∠A = ∠A ( common ) .
∠ADB = ∠ABC [ each equal to 90° ] .
∴ △ADB ∼ △ABC [ By AA-similarity ] .
⇒ AD/AB = AB/AC .
⇒ AB² = AD × AC ............(1) .
In △BDC and △ABC , we have
∠C = ∠C ( common ) .
∠BDC = ∠ABC [ each equal to 90° ] .
∴ △BDC ∼ △ABC [ By AA-similarity ] .
⇒ DC/BC = BC/AC .
⇒ BC² = DC × AC. ............(2) .
Add in equation (1) and (2) , we get
⇒ AB² + BC² = AD × AC + DC × AC .
⇒ AB² + BC² = AC( AD + DC ) .
⇒ AB² + BC² = AC × AC .
⇒AC² = AB² + BC²
Answer 38,
Refer the image.
Answer 39,
1+sin^2A=3sinAcosA , divide both side by cos^A.
sec^2A+tan^2A=3tanA
1+tan^2A+tan^2A=3tanA
2tan^2A-3tanA+1=0
2tan^2A-2tanA-tanA+1=0
2tanA(tanA-1)-1(tanA-1)=0
(tanA-1)(2tanA-1)=0
Either tanA-1=0
tanA=1
proved
Answer 40
The first number greater 10 which when divided by 4 leaves a remainder 3 is 11.
So, the next number will be 11 + 4 = 15
The other numbers in this will be = 15 + 4 = 19 ; 19 + 4 = 23 ; 23 + 4 = 27
So, the AP = 11, 15, 19, 23, 27, 32......
The last term of this AP will be 299.
We have to find the total numbers that lie between 10 and 300, which when divided by 4 leaves a remainder 3.
In fact we have to find 'n' that is the number of terms of the AP.
an = a + (n - 1)d
299 = 11 + (n -1)4
299 = 11 + 4n - 4
299 - 11 + 4 = 4n
292 = 4n
n = 292/4
n = 73
So, total 73 numbers lie between 10 to 300 which when divided by 4 leaves a remainder 3.
