CBSE Class 10 Answered

Ans Q1

Construct a right angled triangle, and mark one angle as 73 and other as 17.

Since cos 73 = m/n, the adjacent side would be m and hypotenuse would be n.

Hence the remaining side, (n²-m²).

cosec73 - cos17= 1/sin73 - cos17 = 1/[(n²-m²)/n] - (n²-m²)/n = (n² - (n²-m²))/n(n²-m²) = m²))/n(n²-m²)

AnsQ2

Since the angle of elevation from A is lower than B, it must be farther than location B.

AB = 50.

AP - BP = AB

AP - BP = 2BP

AP = 3 BP

BP = AB/2 = 50/2 = 25 ..............given.

AP = 75.

tanA = H/75 =         ..........H - height of tower

tanB = H/25

tanB / tanA = 3

tan2A = 3 tanA 

tan(A+A) = 3 tanA

2tanA/(1-tan²A) = 3 tanA

 2 = 3 - 3tan²A

tanA = 1/3

A = 30, hence B = 60.

Height, H = APtan30 = 253 m

Regards,

Team,

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