Here, let O be the origin. The PQ is such that: PMMQ = m1 m2 or, PM→ MQ→ = m1 m2 or, m2PM→ = m1MQ→ or, m2(OM→ - OP→) = m1(OQ→ - OM→) or, m2OM   → - m2p→ = m1q→ – m1OM→ or, .....

Let, PQRS be a rectangle and PR and QS are its diagonals. (i) PR2 = (PR→)2 = (PQ→+ QR→)2             = (SR→+ QR→)2            = SR→2+ 2SR→.QR→ .....

Here, let O be the origin and M divides the line segment AB externally in the ratio of m:n. AMBM = m n or, m BM→ = n AM→ or, m OM→ - OB→ = nOM→ - OA→ or, mOM→ - mb→ = noM→ - na→ or, mOM→ - .....

Here, In the given triangle XYZ, ∡XYZ = 90° and A is the middle point of XZ. To prove: XA = YA = ZA (i)YZ   → = YA→ + AZ→ [By the triangle law of vector .....

Let AB be the height of a tower. Let ∡ACB and ∡ADB are two complementary angles. 

So, ∡ADB = 0 and

Let AB be the height of flagstaff and BC be a tower. Let ∡BDC and ∡ADB be the angles whose values are 30° and 15° respectively. From right angled ∆BCD,      tan θ = pb = BCCD or, tan 30° = BC100 or, 13 = .....

Here,   Let CD = 20 m be the height of a house and AB = x m be the height of the tower. Let ∡ADE = 45° and ∡ACB = 60 ° are the angles of elevation of the top of tower from top and bottom of the house respectively. From right .....

Here, let AB be a vertical pole which is divided by C in the ratio of 2:1. So, AC = 2x and BC = x. Let ∡CDB= 30° be an angle of elevation.  We know that, tan θ = pb    tan 30° = xBD or, 13 = xBD ∴  .....

Here, let, AC= xm be the height of roof of a house. Let AB = 18 m B is the point from the figure, BC = (x -18)m. From right angled ∆BDC       tan 30° = BCDC= x - 18DC or, 13 = x – 18DC ∴    DC = √(3 .....

​​​​​​Here, Let AB the height of a tower CD =20 m be the height of temple. Let, ∡EAD = 60° and ∡DBC =30° be the angle of depression and angle of elevation of the pinnacle of the temple from the top of the tower and bottom of .....

Here,   Let AB be a tower whose height is required to find.  Let, ∡ACB = (90° - θ) and ∡ADB are the angle of elevations. BC = 121 m and BD = 144 m From the right angled ∆ABC,       tan (90° - θ) = .....

Let AB be the height of the towers and BC and BD are the length of shadows. Let ∡ACB = 45° and ∡ADB = 30° be the angle of elevation or heights of sun at two points.  From the right angled ∆ABC,       tan .....

Here, let AB= 21 m be the height of a mountain. Let ∡FAD = 45° and ∡FAC = 60° be the angle of depressions. Let CD = x m be the height of the tower. ∴ ∡FAD = ∡ADE = 45° and ∡FAC = ∡ACB = 60°. From the right angled .....

Here, Let AB = x m be the height of a tower.  Let ∡ADC = 30° and ∡ACB = 45° are the angles of elevation. Let CD = 60 m be the distance between two points. From the right angled ∆ABC,       tan θ = pb or, tan .....

Here,      sin 2θ + cos θ = 0   or, 2sin θ. cos θ + cos θ = 0   or, cos θ (2sin θ+1)=0. Either,     cos θ =0           ⇒  cos .....

Here,      √3 sinθ – cosθ = √2             (0° ≤ θ ≤ 360°) Now,     Coefficient of sinθ2+ (Coefficient of cosθ)2 = 3 2+ .....

Here,       (1 - √3 )tan θ + 1 + √3 = √3 sec2 θ or, (1 - √3 )tan θ + 1 + √3  = √3 (1 + tan2 θ) or, (1 - √3 )tan θ + 1 + √3  = √3  .....

Here,      √3  sinx + cos x = 1   Now,    Coefficient of sinx2+ Coefficient of cosx2 = 3 2+ 12 = 3+1 = 4 = 2. Dividing both side by 2 then, 3 2sin x + 12 cos x = 12 or, cos30° sin x + .....

Here,       2√3 cos2θ = sinθ or, 2√3(1 – sin2θ) = sinθ or, 2√3 - 2√3 sin2θ – sinθ = 0 or, 2√3 sin2θ + sinθ - 2√3 =  or, .....

Here,       2√3 sin2 θ = cos θ   or, 2√3 (1 – cos2 θ) = cos θ   or, 2 √3   - 2√3 cos2 θ   = cos θ   or, 2 √3  cos2 θ .....

Here,       sinA = √3 (1 – cosA)      or, sinA = √3  - √(3 )cosA  or, sinA + √3 cosA = √3 ..…….. (i) Now,     Coefficient of sinA2+ Coefficient .....

Here,       tan θ  =  cotθ  or,tan θ = 1/(tan θ) or, tan2 θ = 1  or, tan θ = ±1.  Since, θ ≤ 90°. So, taking positive value only,     .....

Here,        sinA + sin3A = sin2A  or, 2sin(A+3A2)cos(A-3A2) = sin2A or, 2sin2A cosA – sin2A = 0  or, sin 2A (2cosA – 1) = 0 . Either,                  sin2A = .....

Here,  sin2θ - sin θ + 14 = 0 or, 4sin2 θ - 4sin θ + 1 = 0 or, (2sin θ)2 - 2.2sin θ .1+ 12 = 0 or, (2sin θ - 1)2 = 0 or, 2sin θ - 1 = 0 or, 2sin θ = 1  or, sin θ = 12 or, sin .....

Here, sin2θ - sin θ + 14 = 0 or, 4sin2 θ - 4sin θ + 1 = 0 or, (2sin θ - 1)2 = 0 or, 2sin θ - 1 = 0 or, 2sin θ = 1  or, sin θ = 12 or, 2sin θ = sin30°    [Since, 0° .....

Here,      √3 tan   A+1 = 0       or, √3 tan A  = -1 or,   tanA  = -  1/√3 or,   tanA  = tan (180°-30°)    [Since, 0° ≤A ≤ .....

Here,      3 tan θ - √3  = 0   or, 3 tan θ= √3 or, tan θ = (√3)/3 or, tan θ = 1/(√3) or, tan θ = tan 30°            [since 0° ≤ .....

Here,      √3  tan⁡  θ +3  = 0 or,√3  tan⁡  θ   = -3 or, tan⁡  θ   =  - √3 or, tan θ  =   tan (180°-60°)  .....

Here,       sinA = √(3 )(1 – cosA)  or, sinA = √(3 ) - √(3 )cosA or, sinA + √(3 )cosA = √(3 )……. (i)  Now,      Coefficient of sinA2+ Coefficient of .....

Here,        1 + √3 tanα - sec α or, 1 + 3 sinαcosα - 1cosα = 0 or, cosα+ 3 tanα-1 cosα = 0 or, cosα + √3 sinα - 1 = 0  or, cos α + √3 .....

Here,      √3 tan θ = 3  or, tan θ = √3 or, tan θ = tan60° [∵θ≤  180°] ∴    θ = 60° Thus, the value of  θ  =  60° .....

Here,    sinθ + cosθ = 1 . Now     (coefficient of sinθ)2+ (coefficient of cosθ)2 = 12+ 12 = 2 Dividing given equation by 2 then, 12 sinθ + cosθ 12 = 12 or, sinθcos45° + .....

Here,  TanX = 1 or, TanX = Tan(180 + 45) or, X = .....

Here, cosA = -1/√2 or, cosA = cos(180 - 45) or, A = .....

Here,  4-3 sec2θ = 0 or, 4 = 3 sec2θ  or, 43 = sec2θ or, sec θ = 2/√3 or, sec θ =  sec 30° [∵  θ ≤90°]     ∴ θ =  30°  Thus, value .....

Here,  2√3 cos2θ = sinθ or, 2√3(1 – sin2θ) = sinθ or, 2√3 - 2√3 sin2θ – sinθ = 0 or, 2√3 sin2θ + sinθ - 2√3 =  or, 2√3 sin2θ + .....

Here,        1- tan2θ =  -2        or, 3 = tan2θ or, tan2θ = (± √3) 2  or,  tan⁡θ = ± √3 [∵θ≤  90°] or, tan⁡θ = .....

Here,       sin x - sin2x = 0   or, sin x - 2sinx. cos x = 0   or, sin x (1- 2cos x) = 0 .   Either,       sin x = 0 or, sin x = sin 0° ∴   x = 0°. OR,     .....

Here,        A + B + C = πc or, A + B = πc- C      sin(A + B) = sin(π c - C) sinC       cos(A + B) = cos (πc - C) = -cosC. LHS = cos(B + C – A) + cos(C + A – B) + cos(A + B - .....

Here,         A + B + C = πc        A + B = πc - C. Taking sin and cos on both the sides then, sin (A + B) =  sin (πc-C) =  sin c      cos (A + B) =  cos (πc-C) .....

Here,       A + B + C = π or, A2 + B2 +C2 = π2 or, A2 + B2 =π2-C2 ∴ sinA2 + B2 = cos C2 and ⁡cosA2 + B2 = ⁡sinC2 LHS = sin A - sin B + sin C          = 2⁡cosA+B2⁡sin A-B2++sin .....

Here,           A + B + C = πc          A + B = πc – C  Taking sin and cos on both the sides then, sin (A + B) =  sin (πc-C) =  sin C    cos (A+B) = .....

Here,       A + B =180°- C  ∴  cos (A + B) =  cos (180°-C) =  - cos c      & sin (A + B) =  sin (180°-C) =  sin c. LHS = cos2 A+ cos2 B +2cos A cos B cos .....

Here,        A + B + C = πc  or, A + B = πc - C or, A2+B2=πc2- C2 or, sin( A2+B2)= sin(πc2- C2)= cosC2 or, cos( A2+)= cos (πc2- C2)= sin C2 LHS = cos A + cos B + cos C         = 2 cos .....

Here,        P + Q + R = 180°  or, P2+ Q2+ R2=180° 2 or, P2+ Q2=90° - R2 or, sin (P2+ Q2)=sin (90° -R2) ∴  sin (P2+ Q2)= cos R2 RHS = sin P + sin Q + sin R          = .....

Here,       P + Q + R = π                      or, P + Q = π - R     Operating by sin & cos,       sin (P + Q) =  sin .....

Here,        A + B + C =180° or,  A + B =180°- C . Taking sin and cos on both the sides        sin (A + B) =  sin (180°-C) =  sin C      &   cos (A + .....

Here,        A + B + C = πc or, A + B = πc - C       sin(A + B) = sin(πc) sinC       cos(A +  B) = cos (πc - C) = -cosC. LHS = cos(B + C – A) + cos(C + A – B) .....

Here, LHS  = sin2A + sin 2B +sin2C          = 2 sin(A+B)cos(A-B) + 2sinC cosC          =2sinC cos(A-B)+2sinC cosC          =2sinC (cos(A-b) + cos C)     .....

We know from one of its many properties that, the opposite angles of ABCD are supplementary,   That is    A + C = .....