Question

Answer these .....
also justify ur answer ....
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Answer 1

Answer:

hkdwyehfkuleehshgl the exam of the ceo's observation of consent Angel hsleeyleupsiksttwos the week is required to pay for

Step-by-step explanation:

sgwyueypasysdjgq the week of the original on the week ri sa gkshlwsgld the best of luck in the earth's surface of the

Answer 2

Answer:

Solution:

(i)

(i) The value of tan A is always less than 1.

Answer: False

Proof: In ΔMNC in which ∠N = 90∘,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.

MC2=MN2+NC2

52=32+42

25=9+16

25 = 25

(ii) sec A = 12/5 for some value of angle A

Answer: True

Justification: Let a ΔMNC in which ∠N = 90º,

MC=12k and MB=5k, where k is a positive real number.

By Pythagoras theorem we get,

MC²=MN²+NC²

(12k)²=(5k)²+NC²

NC²+25k2=144k²

NC²=119k²

Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) cos A is the abbreviation used for the cosecant of angle A.

Answer: False

Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.

(iv) cot A is the product of cot and A.

Answer: False

Justification: cot M is not the product of cot and M. It is the cotangent of ∠M.

(v) sin θ = 4/3 for some angle θ.

Answer: False

Justification: sin θ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.