In the diagram, D, B and C are points in the same horizontal plane. AB is a vertical flagpole. A rope is tied from A, the top of the flagpole, to point C. BAC = BCD = θ, BDC = α and AB = 4 units.

Question

Asked on July 3, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-07-03T06:20:08.719Z

QUESTION 8 8.1 Prove that Tt = Pp. Step 1: Draw an altitude from R to the line TP. Let H be the foot of the perpendicular from R to TP. So, RH is the altitude. In right-angled RHT: T = (RH)/(RT) Given RT = p, so: T = (RH)/(p) RH = p T (Equation 1) Step 2: Use the altitude in the other right-angled triangle. In right-angled RHP: P = (RH)/(RP) Given RP = t, so: P = (RH)/(t) RH = t P (Equation 2) Step 3: Equate the expressions for RH and rearrange. From Equation 1 and Equation 2: p T = t P Divide both sides by pt: p Tpt = t Ppt Tt = Pp This proves the identity. Tt = Pp 8.2.1 Show that the distance between B and C is 4 . Step 1: Identify the relevant triangle and its properties. Consider ABC. AB is a vertical flagpole and B, C are points in the horizontal plane, so ABC = 90^. We are given AB = 4 units and BAC = . Step 2: Use a trigonometric ratio to find BC. In right-angled ABC: (BAC) = OppositeAdjacent = (BC)/(AB) Substitute the given values: = (BC)/(4) BC = 4 This shows that the distance between B and C is 4 . BC = 4 8.2.2 Hence, show that DC = 4 (1 + ( )/( )). Step 1: Use the Sine Rule in BCD. We have BC = 4 from 8.2.1. In BCD, we are given BCD = and BDC = . The sum of angles in a triangle is 180^, so DBC = 180^ - ( + ). Using the Sine Rule: (DC)/((D)BC) = (BC)/((B)DC) (DC)/((180^ - ( + ))) = (BC)/( ) Since (180^ - X) = X: (DC)/(( + )) = (BC)/( ) Step 2: Substitute BC and solve for DC. DC = (BC ( + ))/( ) Substitute BC = 4 : DC = (4 ( + ))/( ) Step 3: Expand ( + ) and simplify. DC = (4 ( + ))/( ) DC = 4 ( ( )/( ) + ( )/( ) ) DC = 4 ( + ) Step 4: Manipulate the expression to match the required form. Recall = ( )/( ) and = (1)/( ). DC = 4 ( )/( ) ( (1)/( ) + ) Factor out from the bracket: DC = 4 ( )/( ) ( ( )/( ) + 1 ) DC = 4 ( ( )/( ) + 1 ) DC = 4 ( 1 + ( )/( ) ) This shows the required expression for DC. DC = 4 (1 + ( )/( )) Last free one today — make it count tomorrow, or type /upgrade for unlimited.

Accepted Answer · 2026-07-03T06:20:08.719Z

QUESTION 8 8.1 Prove that Tt = Pp. Step 1: Draw an altitude from R to the line TP. Let H be the foot of the perpendicular from R to TP. So, RH is the altitude. In right-angled RHT: T = (RH)/(RT) Given RT = p, so: T = (RH)/(p) RH = p T (Equation 1) Step 2: Use the altitude in the other right-angled triangle. In right-angled RHP: P = (RH)/(RP) Given RP = t, so: P = (RH)/(t) RH = t P (Equation 2) Step 3: Equate the expressions for RH and rearrange. From Equation 1 and Equation 2: p T = t P Divide both sides by pt: p Tpt = t Ppt Tt = Pp This proves the identity. Tt = Pp 8.2.1 Show that the distance between B and C is 4 . Step 1: Identify the relevant triangle and its properties. Consider ABC. AB is a vertical flagpole and B, C are points in the horizontal plane, so ABC = 90^. We are given AB = 4 units and BAC = . Step 2: Use a trigonometric ratio to find BC. In right-angled ABC: (BAC) = OppositeAdjacent = (BC)/(AB) Substitute the given values: = (BC)/(4) BC = 4 This shows that the distance between B and C is 4 . BC = 4 8.2.2 Hence, show that DC = 4 (1 + ( )/( )). Step 1: Use the Sine Rule in BCD. We have BC = 4 from 8.2.1. In BCD, we are given BCD = and BDC = . The sum of angles in a triangle is 180^, so DBC = 180^ - ( + ). Using the Sine Rule: (DC)/((D)BC) = (BC)/((B)DC) (DC)/((180^ - ( + ))) = (BC)/( ) Since (180^ - X) = X: (DC)/(( + )) = (BC)/( ) Step 2: Substitute BC and solve for DC. DC = (BC ( + ))/( ) Substitute BC = 4 : DC = (4 ( + ))/( ) Step 3: Expand ( + ) and simplify. DC = (4 ( + ))/( ) DC = 4 ( ( )/( ) + ( )/( ) ) DC = 4 ( + ) Step 4: Manipulate the expression to match the required form. Recall = ( )/( ) and = (1)/( ). DC = 4 ( )/( ) ( (1)/( ) + ) Factor out from the bracket: DC = 4 ( )/( ) ( ( )/( ) + 1 ) DC = 4 ( ( )/( ) + 1 ) DC = 4 ( 1 + ( )/( ) ) This shows the required expression for DC. DC = 4 (1 + ( )/( )) Last free one today — make it count tomorrow, or type /upgrade for unlimited.

In the diagram, D, B and C are points in the same horizontal plane. AB is a vertical flagpole. A rope is tied from A, the top of the flagpole, to point C. BAC = BCD = θ, BDC = α and AB = 4 units.

QUESTION 8

8.1 Prove that $\frac{\sin \hat{T}}{t} = \frac{\sin \hat{P}}{p}$ .

8.2.1 Show that the distance between B and C is $4 \tan \theta$ .

8.2.2 Hence, show that $\text{DC} = 4 \sin \theta \left(1 + \frac{\tan \theta}{\tan \alpha}\right)$ .

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In the diagram, D, B and C are points in the same horizontal plane. AB is a vertical flagpole. A rope is tied from A, the top of the flagpole, to point C. BAC = BCD = θ, BDC = α and AB = 4 units.

QUESTION 8

8.1 Prove that $\frac{\sin \hat{T}}{t} = \frac{\sin \hat{P}}{p}$ .

8.2.1 Show that the distance between B and C is $4 \tan \theta$ .

8.2.2 Hence, show that $\text{DC} = 4 \sin \theta \left(1 + \frac{\tan \theta}{\tan \alpha}\right)$ .

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In the diagram, D, B and C are points in the same horizontal plane. AB is a vertical flagpole. A rope is tied from A, the top of the flagpole, to point C. BAC = BCD = θ, BDC = α and AB = 4 units.

QUESTION 8

8.1 Prove that sin⁡T^t=sin⁡P^p\frac{\sin \hat{T}}{t} = \frac{\sin \hat{P}}{p}tsinT^​=psinP^​.

8.2.1 Show that the distance between B and C is 4tan⁡θ4 \tan \theta4tanθ.

8.2.2 Hence, show that DC=4sin⁡θ(1+tan⁡θtan⁡α)\text{DC} = 4 \sin \theta \left(1 + \frac{\tan \theta}{\tan \alpha}\right)DC=4sinθ(1+tanαtanθ​).

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In the diagram, D, B and C are points in the same horizontal plane. AB is a vertical flagpole. A rope is tied from A, the top of the flagpole, to point C. BAC = BCD = θ, BDC = α and AB = 4 units.

QUESTION 8

8.1 Prove that sin⁡T^t=sin⁡P^p\frac{\sin \hat{T}}{t} = \frac{\sin \hat{P}}{p}tsinT^​=psinP^​.

8.2.1 Show that the distance between B and C is 4tan⁡θ4 \tan \theta4tanθ.

8.2.2 Hence, show that DC=4sin⁡θ(1+tan⁡θtan⁡α)\text{DC} = 4 \sin \theta \left(1 + \frac{\tan \theta}{\tan \alpha}\right)DC=4sinθ(1+tanαtanθ​).

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8.1 Prove that $\frac{\sin \hat{T}}{t} = \frac{\sin \hat{P}}{p}$ .

8.2.1 Show that the distance between B and C is $4 \tan \theta$ .

8.2.2 Hence, show that $\text{DC} = 4 \sin \theta \left(1 + \frac{\tan \theta}{\tan \alpha}\right)$ .

8.1 Prove that $\frac{\sin \hat{T}}{t} = \frac{\sin \hat{P}}{p}$ .

8.2.1 Show that the distance between B and C is $4 \tan \theta$ .

8.2.2 Hence, show that $\text{DC} = 4 \sin \theta \left(1 + \frac{\tan \theta}{\tan \alpha}\right)$ .