Given that: tan(A + B) = √3 ⇒ tan (A+B) = tan 60° [∵ tan 60° = √3] ⇒ A+B= 60° ...(i) Given that: tan (A - B) = 1/√3 ⇒ tan(A-B) = tan 30° [∵ tan 30° = 1/√3] ⇒ A-B = 30° ...(ii) Solving the equations (i) and (ii), we get 2A = 90° ⇒ A = 45° From equation (1), we get 45°+B = 60° ⇒ B = 15° Hence, A = 45°Read more
Given that: tan(A + B) = √3
⇒ tan (A+B) = tan 60° [∵ tan 60° = √3]
⇒ A+B= 60° …(i)
Given that: tan (A – B) = 1/√3
⇒ tan(A-B) = tan 30° [∵ tan 30° = 1/√3]
⇒ A-B = 30° …(ii)
Solving the equations (i) and (ii), we get
2A = 90° ⇒ A = 45°
From equation (1), we get
45°+B = 60° ⇒ B = 15°
Hence, A = 45° and B = 15°
See this video Solution for better understanding 🙌😀
Given that: cot 0 = 7/8 Let cot 0=where k is a real number. In △ABC, by Pythagoras theorem, we have AC² = BC²+AB² = (8k)²+(7k)² = 64k²+49k² = 113k² ⇒ AC = √(113k²) = √(113k) (i) ((1+sin θ)(1-sin 9))/((1+cos θ)(1-cos θ)) = ((1+ 8/√113)(1- 8/√113))/((1+ 7√113)(1- 7/√113)) = (1-(8/√113)²)/(1-(7/√113)²)Read more
Given that: cot 0 = 7/8
Let cot 0=where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (8k)²+(7k)²
= 64k²+49k²
= 113k²
⇒ AC = √(113k²) = √(113k)
(i) ((1+sin θ)(1-sin 9))/((1+cos θ)(1-cos θ))
= ((1+ 8/√113)(1- 8/√113))/((1+ 7√113)(1- 7/√113))
= (1-(8/√113)²)/(1-(7/√113)²) = (1- 64/113)/(1- 49/113) = ((113-64)/113)/((113-49)/113) = 49/64
(ii) cot² θ = (cot θ)² = (7/8)² = 49/64
Here is the Video explanation of the above question😀👇
Given that: sec 0 = 13/12 Let sec 0 = 13k/12k, where k is a real number. In △ABC, by Pythagoras theorem, we have BC² = AC² - AB² = (13k)² - (12k)² = 169k² 144k² = 25k² ⇒ BC = √(25k²) = 5k Hence, sin 0 = BC/AC = 5k/13k = 5/13 cos 0 = AB/AC = 12k/13k = 12/13 tan 0 = BC/AB = 5k/12k = 5/12 cose 0 = AC/BRead more
Given that: sec 0 = 13/12
Let sec 0 = 13k/12k, where k is a real number.
In △ABC, by Pythagoras theorem, we have
BC² = AC² – AB²
= (13k)² – (12k)²
= 169k² 144k²
= 25k²
⇒ BC = √(25k²) = 5k
Hence, sin 0 = BC/AC = 5k/13k = 5/13
cos 0 = AB/AC = 12k/13k = 12/13
tan 0 = BC/AB = 5k/12k = 5/12
cose 0 = AC/BC = 13k/5k = 13/5 and cot 0 = AB/BC = 12k/5k = 12/5
See the below video for better explanation of the above question👇😀
Given that: cos A= cos B cos A = cos B ⇒ AP/AQ = BC/BD ⇒ AP/BC = AQ/BD Let AP/BC = AQ/BC = k ...(i) Therefore, AP = k(BC) and AQ = k(BD) Now, In △APQ and △BCD PQ/CD = √((AQ²-AP²))/(√(BD²-BC²)) = (√((k.BD)²-(k.BC)²))/(√(BD²-BC²)) = (k√(BD²-BC²))/(√(BD²-BC²)) = k ...(ii) From the equation (i) and (ii)Read more
Given that: cos A= cos B
cos A = cos B
⇒ AP/AQ = BC/BD
⇒ AP/BC = AQ/BD
Let AP/BC = AQ/BC = k …(i)
Therefore, AP = k(BC) and AQ = k(BD)
Now, In △APQ and △BCD
PQ/CD = √((AQ²-AP²))/(√(BD²-BC²))
= (√((k.BD)²-(k.BC)²))/(√(BD²-BC²))
= (k√(BD²-BC²))/(√(BD²-BC²)) = k …(ii)
From the equation (i) and (ii), we get
AP/BC = AQ/BD = PQ/CD
So, △APQ ∼ △BCD [SSS similarity criteria]
Hence, ∠A = ∠B
(i) The median through A meet BC at D. So, D is the mid-point of BC. Therefore Coordinates of D = ((6+1)/2, (5+4)/2) = (7/2, 9/2) (ii) Point P lies on AD, such that AP:PD = 2:1. Therefore (1.4) Coordinates of P = ((2×7/2+1×4)/(2+1), (2×9/2+1×2)/(2+1)) (iii) The median through B meets AC at E. So, ERead more
(i) The median through A meet BC at D. So, D is the mid-point of BC. Therefore Coordinates of D = ((6+1)/2, (5+4)/2) = (7/2, 9/2)
(ii) Point P lies on AD, such that AP:PD = 2:1. Therefore (1.4) Coordinates of P = ((2×7/2+1×4)/(2+1), (2×9/2+1×2)/(2+1))
(iii) The median through B meets AC at E. So, E is the mid-point of AC. Therefore, Coordinates of E = ((4+1)/2, (2+4)/2) = (5/2, 3)
(iv) Point Q lies on AE, such that AQ:QE = 2:1.Therefore Coordinates of Q = ((2×5/2+1×6)/(2+1), (2×3+1×5)/(2+1)) = (11/3, 11/3)
The median through C mects AB at F. So, F is the mid-point of AB. Therefore Coordinates of F = ((4+6)/2, (2+5)/2) = (5, 7/2)
Point R lies on CF, such that CR:RF = 2:1.
(v) Therefore, the coordinates of R = ((2×5+1×1)/(2+1), (2×7/2+1×4)/(2+1)) = (11/3, 11/3)
(vi) The coordinates of P, Q and R is same.
(vii)Points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of triangle ABC. Median through A meets BC at D. So, D is the mid-point of BC. Therefore, The coordinates of D = ((x₂+x₃)/2, (y₂+y₃)/2)
Let O be the centroid of the triangle. Point O lies on AD such that A0:0D = 2:1.
Therefore, the coordinates of point O
((2×(x₂+x₃)/2 +1×x₁)/(2+1), (2×(y₂+y₃)/2 +1×y₁)/(2+1)) = ((x₁+x₂+x₃)/2, (y₁+y₂+y₃)/2)
Here you can see the video explanation of this question ✋
If tan (A + B) = √3 and tan (A – B) = 1/√3 ; 0° B, find A and B.
Given that: tan(A + B) = √3 ⇒ tan (A+B) = tan 60° [∵ tan 60° = √3] ⇒ A+B= 60° ...(i) Given that: tan (A - B) = 1/√3 ⇒ tan(A-B) = tan 30° [∵ tan 30° = 1/√3] ⇒ A-B = 30° ...(ii) Solving the equations (i) and (ii), we get 2A = 90° ⇒ A = 45° From equation (1), we get 45°+B = 60° ⇒ B = 15° Hence, A = 45°Read more
Given that: tan(A + B) = √3
⇒ tan (A+B) = tan 60° [∵ tan 60° = √3]
⇒ A+B= 60° …(i)
Given that: tan (A – B) = 1/√3
⇒ tan(A-B) = tan 30° [∵ tan 30° = 1/√3]
⇒ A-B = 30° …(ii)
Solving the equations (i) and (ii), we get
2A = 90° ⇒ A = 45°
From equation (1), we get
45°+B = 60° ⇒ B = 15°
Hence, A = 45° and B = 15°
See this video Solution for better understanding 🙌😀
See lessIf cot θ = 7/8, evaluate: (i) (1-sin) (1+sinθ) / (1-cosθ) (1+cosθ) (ii) cot²θ
Given that: cot 0 = 7/8 Let cot 0=where k is a real number. In △ABC, by Pythagoras theorem, we have AC² = BC²+AB² = (8k)²+(7k)² = 64k²+49k² = 113k² ⇒ AC = √(113k²) = √(113k) (i) ((1+sin θ)(1-sin 9))/((1+cos θ)(1-cos θ)) = ((1+ 8/√113)(1- 8/√113))/((1+ 7√113)(1- 7/√113)) = (1-(8/√113)²)/(1-(7/√113)²)Read more
Given that: cot 0 = 7/8
Let cot 0=where k is a real number.
In △ABC, by Pythagoras theorem, we have
AC² = BC²+AB²
= (8k)²+(7k)²
= 64k²+49k²
= 113k²
⇒ AC = √(113k²) = √(113k)
(i) ((1+sin θ)(1-sin 9))/((1+cos θ)(1-cos θ))
= ((1+ 8/√113)(1- 8/√113))/((1+ 7√113)(1- 7/√113))
= (1-(8/√113)²)/(1-(7/√113)²) = (1- 64/113)/(1- 49/113) = ((113-64)/113)/((113-49)/113) = 49/64
(ii) cot² θ = (cot θ)² = (7/8)² = 49/64
Here is the Video explanation of the above question😀👇
See lessGiven sec θ = 13/12, Calculate all other trigonometric ratios.
Given that: sec 0 = 13/12 Let sec 0 = 13k/12k, where k is a real number. In △ABC, by Pythagoras theorem, we have BC² = AC² - AB² = (13k)² - (12k)² = 169k² 144k² = 25k² ⇒ BC = √(25k²) = 5k Hence, sin 0 = BC/AC = 5k/13k = 5/13 cos 0 = AB/AC = 12k/13k = 12/13 tan 0 = BC/AB = 5k/12k = 5/12 cose 0 = AC/BRead more
Given that: sec 0 = 13/12
Let sec 0 = 13k/12k, where k is a real number.
In △ABC, by Pythagoras theorem, we have
BC² = AC² – AB²
= (13k)² – (12k)²
= 169k² 144k²
= 25k²
⇒ BC = √(25k²) = 5k
Hence, sin 0 = BC/AC = 5k/13k = 5/13
cos 0 = AB/AC = 12k/13k = 12/13
tan 0 = BC/AB = 5k/12k = 5/12
cose 0 = AC/BC = 13k/5k = 13/5 and cot 0 = AB/BC = 12k/5k = 12/5
See the below video for better explanation of the above question👇😀
See lessIf angle A and angle B are acute angles such that cos A = cos B, then show that angle A = angle B.
Given that: cos A= cos B cos A = cos B ⇒ AP/AQ = BC/BD ⇒ AP/BC = AQ/BD Let AP/BC = AQ/BC = k ...(i) Therefore, AP = k(BC) and AQ = k(BD) Now, In △APQ and △BCD PQ/CD = √((AQ²-AP²))/(√(BD²-BC²)) = (√((k.BD)²-(k.BC)²))/(√(BD²-BC²)) = (k√(BD²-BC²))/(√(BD²-BC²)) = k ...(ii) From the equation (i) and (ii)Read more
Given that: cos A= cos B
cos A = cos B
⇒ AP/AQ = BC/BD
⇒ AP/BC = AQ/BD
Let AP/BC = AQ/BC = k …(i)
Therefore, AP = k(BC) and AQ = k(BD)
Now, In △APQ and △BCD
PQ/CD = √((AQ²-AP²))/(√(BD²-BC²))
= (√((k.BD)²-(k.BC)²))/(√(BD²-BC²))
= (k√(BD²-BC²))/(√(BD²-BC²)) = k …(ii)
From the equation (i) and (ii), we get
AP/BC = AQ/BD = PQ/CD
So, △APQ ∼ △BCD [SSS similarity criteria]
Hence, ∠A = ∠B
Video explanation of this question👌🤗
See lessLet A(4, 2), B(6, 5) and C(1, 4) be the vertices of triangle ABC.
(i) The median through A meet BC at D. So, D is the mid-point of BC. Therefore Coordinates of D = ((6+1)/2, (5+4)/2) = (7/2, 9/2) (ii) Point P lies on AD, such that AP:PD = 2:1. Therefore (1.4) Coordinates of P = ((2×7/2+1×4)/(2+1), (2×9/2+1×2)/(2+1)) (iii) The median through B meets AC at E. So, ERead more
(i) The median through A meet BC at D. So, D is the mid-point of BC. Therefore Coordinates of D = ((6+1)/2, (5+4)/2) = (7/2, 9/2)
(ii) Point P lies on AD, such that AP:PD = 2:1. Therefore (1.4) Coordinates of P = ((2×7/2+1×4)/(2+1), (2×9/2+1×2)/(2+1))
(iii) The median through B meets AC at E. So, E is the mid-point of AC. Therefore, Coordinates of E = ((4+1)/2, (2+4)/2) = (5/2, 3)
(iv) Point Q lies on AE, such that AQ:QE = 2:1.Therefore Coordinates of Q = ((2×5/2+1×6)/(2+1), (2×3+1×5)/(2+1)) = (11/3, 11/3)
The median through C mects AB at F. So, F is the mid-point of AB. Therefore Coordinates of F = ((4+6)/2, (2+5)/2) = (5, 7/2)
Point R lies on CF, such that CR:RF = 2:1.
(v) Therefore, the coordinates of R = ((2×5+1×1)/(2+1), (2×7/2+1×4)/(2+1)) = (11/3, 11/3)
(vi) The coordinates of P, Q and R is same.
(vii)Points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of triangle ABC. Median through A meets BC at D. So, D is the mid-point of BC. Therefore, The coordinates of D = ((x₂+x₃)/2, (y₂+y₃)/2)
Let O be the centroid of the triangle. Point O lies on AD such that A0:0D = 2:1.
Therefore, the coordinates of point O
((2×(x₂+x₃)/2 +1×x₁)/(2+1), (2×(y₂+y₃)/2 +1×y₁)/(2+1)) = ((x₁+x₂+x₃)/2, (y₁+y₂+y₃)/2)
Here you can see the video explanation of this question ✋
See less