Question

tan - 7
Q.2. B) Solve the following questions. (Any-4)
08
1) Prove that. cos( 1+tan-0) - 1
2) A person is standing at a distance of 80m from a church looking at its top.
The angle of elevation is 45°. Find the height of church
3) Find the length of diagonal of a rectangle having sides llem and 60cm.
4) In ARST, ZS - 90", ZT - 30°, RT = 12em then find RS and ST.
5) Areas of two similar triangles are 225 sq em and 81 sqem. If a side of
smaller triangle is 2 em then find corresponding side of bigger triangle
Q.3. A) Complete the following activities.(Any-1)
1) For finding AB and BC with the help of information givn in figure complete
the following activity
AB=BC
.. ZBAC =
AB=BC
IX AC
XV&
x272
B​

Answer 1

Answer:

1) cos ²∅(1 + tan ²∅) = 1

Proof:

cos ²∅ x sec ²∅ [1+ tan ²∅ =  sec²∅]

cos ²∅ x 1/cos ²∅

1 = 1

2) Let the height of the church be h

tan 45° = h/80

1 = h/80

∴ h = 80 m

3) In a rectangle all angles are 90°,

So, by Pythagoras Theorem,

Diagonal ² = 11 ² + 60 ²

Diagonal  = √(121 + 3600)

Diagonal = √(3721)

Diagonal = 61 cm

4) As ∠T = 30°,

sin 30° = RS/RT

1/2 = RS/12

RS = 6 cm

cos 30° = ST/RT

√(3)/2 = ST/12

ST = 6√(3) cm

5) The ratio of the areas of similar triangles is equal to the ratio of the squares of its sides.

Let the similar side be y

225/81 = y²/12²

√[(225 x 144)/81] = y

y = (25 x 12)/9

y = 100/3 or 33.33 cm

3A)  

AB = BC = 2 u

∠BAC = 45°

AB = BC = 1/√(2) x AC

1/√(2) x 2√(2)

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