Question 1:
In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine
(i) sin A, cos A
(ii) sin C, cos C
Applying Pythagoras theorem for ΔABC, we obtain
AC2 = AB2 + BC2
= (24 cm)2 + (7 cm)2
= (576 + 49) cm2
= 625 cm2
∴ AC =
cm = 25 cm
(i) sin A =
cos A =
(ii)
sin C =
cos C =
AC2 = AB2 + BC2
= (24 cm)2 + (7 cm)2
= (576 + 49) cm2
= 625 cm2
∴ AC =
cm = 25 cm(i) sin A =
cos A =
(ii)
sin C =
cos C =
Question 2:
In the given figure find tan P − cot R
Applying Pythagoras theorem for ΔPQR, we obtain
PR2 = PQ2 + QR2
(13 cm)2 = (12 cm)2 + QR2
169 cm2 = 144 cm2 + QR2
25 cm2 = QR2
QR = 5 cm
tan P − cot R =
PR2 = PQ2 + QR2
(13 cm)2 = (12 cm)2 + QR2
169 cm2 = 144 cm2 + QR2
25 cm2 = QR2
QR = 5 cm
tan P − cot R =
Question 3:
If sin A =
, calculate cos A and tan A.
, calculate cos A and tan A.
Let ΔABC be a right-angled triangle, right-angled at point B.
Given that,
Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
(4k)2 = AB2 + (3k)2
16k 2 − 9k 2 = AB2
7k 2 = AB2
AB =
Given that,
Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
(4k)2 = AB2 + (3k)2
16k 2 − 9k 2 = AB2
7k 2 = AB2
AB =
Question 4:
Given 15 cot A = 8. Find sin A and sec A
Consider a right-angled triangle, right-angled at B.
It is given that,
cot A =
Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
= (8k)2 + (15k)2
= 64k2 + 225k2
= 289k2
AC = 17k
It is given that,
cot A =
Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
= (8k)2 + (15k)2
= 64k2 + 225k2
= 289k2
AC = 17k
Question 5:
Given sec θ =
, calculate all other trigonometric ratios.
, calculate all other trigonometric ratios.
Consider a right-angle triangle ΔABC, right-angled at point B.
If AC is 13k, AB will be 12k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
(AC)2 = (AB)2 + (BC)2
(13k)2 = (12k)2 + (BC)2
169k2 = 144k2 + BC2
25k2 = BC2
BC = 5k
If AC is 13k, AB will be 12k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
(AC)2 = (AB)2 + (BC)2
(13k)2 = (12k)2 + (BC)2
169k2 = 144k2 + BC2
25k2 = BC2
BC = 5k
Question 6:
If ∠A and ∠B are acute angles such that cos A = cos B, then show that
∠A = ∠B.
Let us consider a triangle ABC in which CD ⊥ AB.
It is given that
cos A = cos B
We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.
From equation (1), we obtain
By using the converse of B.P.T,
CD||BP
⇒∠ACD = ∠CPB (Corresponding angles) … (3)
And, ∠BCD = ∠CBP (Alternate interior angles) … (4)
By construction, we have BC = CP.
∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)
From equations (3), (4), and (5), we obtain
∠ACD = ∠BCD … (6)
In ΔCAD and ΔCBD,
∠ACD = ∠BCD [Using equation (6)]
∠CDA = ∠CDB [Both 90°]
Therefore, the remaining angles should be equal.
∴∠CAD = ∠CBD
⇒ ∠A = ∠B
Alternatively,
Let us consider a triangle ABC in which CD ⊥ AB.
It is given that,
cos A = cos B
Let
⇒ AD = k BD … (1)
And, AC = k BC … (2)
Using Pythagoras theorem for triangles CAD and CBD, we obtain
CD2 = AC2 − AD2 … (3)
And, CD2 = BC2 − BD2 … (4)
From equations (3) and (4), we obtain
AC2 − AD2 = BC2 − BD2
⇒ (k BC)2 − (k BD)2 = BC2 − BD2
⇒ k2 (BC2 − BD2) = BC2 − BD2
⇒ k2 = 1
⇒ k = 1
Putting this value in equation (2), we obtain
AC = BC
⇒ ∠A = ∠B(Angles opposite to equal sides of a triangle)
It is given that
cos A = cos B
… (1)We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP.
From equation (1), we obtain
By using the converse of B.P.T,
CD||BP
⇒∠ACD = ∠CPB (Corresponding angles) … (3)
And, ∠BCD = ∠CBP (Alternate interior angles) … (4)
By construction, we have BC = CP.
∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)
From equations (3), (4), and (5), we obtain
∠ACD = ∠BCD … (6)
In ΔCAD and ΔCBD,
∠ACD = ∠BCD [Using equation (6)]
∠CDA = ∠CDB [Both 90°]
Therefore, the remaining angles should be equal.
∴∠CAD = ∠CBD
⇒ ∠A = ∠B
Alternatively,
Let us consider a triangle ABC in which CD ⊥ AB.
It is given that,
cos A = cos B
Let
And, AC = k BC … (2)
Using Pythagoras theorem for triangles CAD and CBD, we obtain
CD2 = AC2 − AD2 … (3)
And, CD2 = BC2 − BD2 … (4)
From equations (3) and (4), we obtain
AC2 − AD2 = BC2 − BD2
⇒ (k BC)2 − (k BD)2 = BC2 − BD2
⇒ k2 (BC2 − BD2) = BC2 − BD2
⇒ k2 = 1
⇒ k = 1
Putting this value in equation (2), we obtain
AC = BC
⇒ ∠A = ∠B(Angles opposite to equal sides of a triangle)
Question 7:
If cot θ =
, evaluate
, evaluate
(i)
(ii) cot2 θ
(ii) cot2 θ
Let us consider a right triangle ABC, right-angled at point B.
If BC is 7k, then AB will be 8k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
= (8k)2 + (7k)2
= 64k2 + 49k2
= 113k2
AC =
(ii) cot2 θ = (cot θ)2 =
=
If BC is 7k, then AB will be 8k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
= (8k)2 + (7k)2
= 64k2 + 49k2
= 113k2
AC =
(ii) cot2 θ = (cot θ)2 =
Question 8:
If 3 cot A = 4, Check whether
It is given that 3cot A = 4
Or, cot A =
Consider a right triangle ABC, right-angled at point B.
If AB is 4k, then BC will be 3k, where k is a positive integer.
In ΔABC,
(AC)2 = (AB)2 + (BC)2
= (4k)2 + (3k)2
= 16k2 + 9k2
= 25k2
AC = 5k
cos2 A − sin2 A =
Or, cot A =
If AB is 4k, then BC will be 3k, where k is a positive integer.
In ΔABC,
(AC)2 = (AB)2 + (BC)2
= (4k)2 + (3k)2
= 16k2 + 9k2
= 25k2
AC = 5k
cos2 A − sin2 A =
Question 9:
In ΔABC, right angled at B. If
find the value of
find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C − sin A sin C
If BC is k, then AB will be, k is a positive integer.
In ΔABC,
AC2 = AB2 + BC2
=
∴ AC = 2k
(i) sin A cos C + cos A sin C
(ii) cos A cos C − sin A sin C
Question 10:
In ΔPQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Given that, PR + QR = 25
PQ = 5
Let PR be x.
Therefore, QR = 25 − x
Applying Pythagoras theorem in ΔPQR, we obtain
PR2 = PQ2 + QR2
x2 = (5)2 + (25 − x)2
x2 = 25 + 625 + x2 − 50x
50x = 650
x = 13
Therefore, PR = 13 cm
QR = (25 − 13) cm = 12 cm
PQ = 5
Let PR be x.
Therefore, QR = 25 − x
Applying Pythagoras theorem in ΔPQR, we obtain
PR2 = PQ2 + QR2
x2 = (5)2 + (25 − x)2
x2 = 25 + 625 + x2 − 50x
50x = 650
x = 13
Therefore, PR = 13 cm
QR = (25 − 13) cm = 12 cm
Question 11:
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A =
for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A
(v) sin θ =
, for some angle θ
(i) Consider a ΔABC, right-angled at B.
But
> 1
∴tan A > 1
So, tan A < 1 is not always true.
Hence, the given statement is false.
(ii)
Let AC be 12k, AB will be 5k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
(12k)2 = (5k)2 + BC2
144k2 = 25k2 + BC2
BC2 = 119k2
BC = 10.9k
It can be observed that for given two sides AC = 12k and AB = 5k,
BC should be such that,
AC − AB < BC < AC + AB
12k − 5k < BC < 12k + 5k
7k < BC < 17 k
However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible.
Hence, the given statement is true.
(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A.
Hence, the given statement is false.
(iv) cot A is not the product of cot and A. It is the cotangent of ∠A.
Hence, the given statement is false.
(v) sin θ =
We know that in a right-angled triangle,
In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.
Hence, the given statement is false
But
∴tan A > 1
So, tan A < 1 is not always true.
Hence, the given statement is false.
(ii)
Let AC be 12k, AB will be 5k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
(12k)2 = (5k)2 + BC2
144k2 = 25k2 + BC2
BC2 = 119k2
BC = 10.9k
It can be observed that for given two sides AC = 12k and AB = 5k,
BC should be such that,
AC − AB < BC < AC + AB
12k − 5k < BC < 12k + 5k
7k < BC < 17 k
However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible.
Hence, the given statement is true.
(iii) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A.
Hence, the given statement is false.
(iv) cot A is not the product of cot and A. It is the cotangent of ∠A.
Hence, the given statement is false.
(v) sin θ =
In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.
Hence, the given statement is false