Trigonometric Ratios NCERT Exercise 8.1 part two Class Ten Mathematics

Trigonometry

Exercise 8.1 Part 2

Question – 7 - If `text(cot θ)=7/8`, evaluate the following:

Solution: `text(cot θ)=7/8=b/p`

This means, b = 7 and p = 8.

We can calculate h by using Pythagoras theorem;

`h^2 = p^2 + b^2`

Or, `h^2 = 8^2 + 7^2`

`= 64 + 49 = 113`

Or, `h = sqrt(113)`


(a) `((1+text(sin θ))(1-text(sin θ)))/((1+text(cos θ))(1-text(cos θ))`

Solution:

`((1+text(sin θ))(1-text(sin θ)))/((1+text(cos θ))(1-text(cos θ))`

`=(1-text(sin)^2θ)/(1-text(cos)^2θ)`

`=(1-(8/sqrt(113))^2)/(1-(7/sqrt(113))^2)`

`=(1-(64)/(113))/(1-(49)/(113))`

`=(113-64)/(113-49)=(49)/(64)`

(b) cot2 θ

Solution:

`text(cot)^2θ=(7/8)^2=(49)/(64)`

Question – 8 - If 3 cot A = 4, check whether `(1-text(tan)^2A)/(1+text(tan)^2A)=text(cos)^2A-text(sin)^2A` or not.

Solution: 3 cot A = 4 means cot A = 4/3 = b/p

Hence, p = 3 and b = 4.

We can calculate h by using Pythagoras theorem;

`h^2 = p^2 + b^2`

Or, `h^2 = 3^2 + 4^2`

`= 9 + 16 = 25`

Or, `h = 5`

Now; the equation can be checked as follows:

`(1-text(tan)^2A)/(1+text(tan)^2A)`

`=(1-(3/4)^2)/(1+(3/4)^2`

`=(1-(9)/(16))/(1+(9)/(16))=(7)/(25)`

RHS:

`text(cos)^2A-text(sin)^2A`

`=(4/5)^2-(3/5)^2`

`=(16)/(25)-(9)/(25)=(7)/(25)`

It is clear that LHS = RHS.


Question – 9 - In triangle ABC, right-angled at B, if `text(tan A)=1/sqrt3` find the value of:

Solution:

`text(tan A)=1/sqrt3=p/b`

This means, p = 1 and `b = sqrt3`

We can calculate h by using Pythagoras theorem;

`h^2 = p^2 + b^2`

Or, `h^2 = 1^2 + (sqrt3)^2`

`= 1 + 3 = 4`

Or, `h = sqrt2=2`

(a) Sin A Cos C + Cos A Sin C

Solution: Sin A Cos C + Cos A Sin C

`=1/2xx1/2+sqrt3/2xxsqrt3/2`

`=1/4+3/4=1`

(Note: side opposite to angle A = side adjacent to angle C and side adjacent to angle A = side opposite to angle C)

(b) Cos A Cos C – Sin A Sin C

Solution: Cos A Cos C – Sin A Sin C

`=sqrt3/2xx1/2-1/2xxsqrt3/2=0`


Question – 10 - In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

10 trigonometry exericse solution

Solution: Given; PR + QR = 25 cm and PQ = 5 cm.

Hence, `PR = 25 – QR`

We can calculate PR and QR by using Pythagoras theorem;

`PQ^2 = PR^2 – OR^2`

Or, `5^2 = (25 – OR)^2 – OR^2`

Or, `25 = 625 + OR^2 – 50OR – OR^2`

Or, `25 = 625 – 50QR`

Or, `50 QR = 625 – 25 = 600`

Or, `QR = 12`

Hence, `PR = 25 – 12 = 13`

Now;

`text(sin P)=(QR)/(PR)=(12)/(13)`

`text(cos P)=(QP)/(PR)=(5)/(13)`

`text(tan P)=(QR)/(QP)=(12)/(5)`

Question – 11 - State whether the following are true or false. Justify your answer.

  1. The value of tan A is always less than 1.

    Solution: False; value of tan begins from zero and goes on to become more than 1.
  2. sec A = 12/5 for some value of angle A.

    Solution: True, value of cos is always more than 1.
  3. cos A is the abbreviation used for the cosecant of angle A.

    Solution: False, cos is the abbreviation of cosine.
  4. cot A is the product of cot and A.

    Solution: False, cot A means cotangent of angle A.
  5. sin A = 4/3 for some angle A.

    Solution: False, value of sin is less than or equal to 1, while this value is more than 1.


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