Successive Discount 1 Concepts

1. Elevation & Depression Angle Shifts

• Formula: For a vertical object of height (h):

$$\tan\theta=\frac{h}{d} \quad\Rightarrow\quad h=d\tan\theta$$

If the observer moves away by (x):

$$d_2=d_1+x$$

and:

$$d_1\tan\theta_1=d_2\tan\theta_2$$

• Example: The angle of elevation of a tower changes from (60^\circ) to (30^\circ) when the observer moves 20 m away. Find the tower height. Solution:

          T
          |\
        h | \
          |  \
          |   \
          |    \
          B-----A-----------C
             d     20 m
           60°     30°

Let (BA=d). Then (BC=d+20).

From A:

$$h=d\tan60^\circ=d\sqrt3$$

From C:

$$h=(d+20)\tan30^\circ =\frac{d+20}{\sqrt3}$$

Equating:

$$d\sqrt3=\frac{d+20}{\sqrt3}$$ $$3d=d+20$$ $$d=10$$

Therefore:

$$h=10\sqrt3$$ $$\boxed{h=10\sqrt3\text{ m}}$$

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2. Angle of Depression

• Formula: The angle of depression from the top of an object equals the angle of elevation from the observer:

$$\text{Angle of depression}=\text{Angle of elevation}$$

Hence:

$$\tan\theta=\frac{\text{vertical height difference}}{\text{horizontal distance}}$$

• Example: From the top of a 20 m building, the angle of depression of a car is (45^\circ). Find the horizontal distance of the car from the building. Solution:

          A
          |\
      20m | \
          |  \
          |   \ 45°
          |    \
          B-----C
             d

$$\tan45^\circ=\frac{20}{d}$$ $$1=\frac{20}{d}$$ $$\boxed{d=20\text{ m}}$$

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3. Multi-Point Shadow Analysis

• Formula: For an object of height (h) casting a shadow of length (s):

$$\tan\theta=\frac{h}{s}$$

where (\theta) is the sun’s angle of elevation.

• Example: A 6 m pole casts a (2\sqrt3) m shadow. Find the sun’s elevation angle. Solution:

       Pole
        |
      6 |\
        | \
        |  \
        |___\ 
          2√3

$$\tan\theta=\frac{6}{2\sqrt3} =\sqrt3$$

Since:

$$\tan60^\circ=\sqrt3$$ $$\boxed{\theta=60^\circ}$$

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4. Shadow Length from Sun’s Elevation

• Formula:

$$s=\frac{h}{\tan\theta}=h\cot\theta$$

• Example: A 10 m pole stands vertically. If the sun’s elevation is (45^\circ), find its shadow length. Solution:

       Pole
        |
      10|\
        | \
        |  \
        |___\
           s

$$\tan45^\circ=\frac{10}{s}$$ $$1=\frac{10}{s}$$ $$\boxed{s=10\text{ m}}$$

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5. Shadow Length Change

• Formula:

$$s_1=h\cot\theta_1,\qquad s_2=h\cot\theta_2$$

Therefore:

$$\frac{s_1}{s_2}=\frac{\cot\theta_1}{\cot\theta_2}$$

• Example: A pole casts a 10 m shadow when the sun’s elevation is (45^\circ). Find the shadow length when the elevation becomes (30^\circ). Solution:

First find height:

$$h=10\tan45^\circ=10$$

At (30^\circ):

$$s=\frac{10}{\tan30^\circ}$$ $$=\frac{10}{1/\sqrt3} =10\sqrt3$$ $$\boxed{10\sqrt3\text{ m}}$$

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6. Dual-Tower & River Width Calculations

• Formula: If two towers of heights (h_1,h_2) are observed from a point at horizontal distances (d_1,d_2):

$$\tan\theta=\frac{h_1}{d_1}, \qquad \tan\phi=\frac{h_2}{d_2}$$

If the observation point is midway between the towers, (d_1=d_2=d).

• Example: Two towers are 30 m and 50 m high. From the midpoint between them, their angles of elevation are complementary. Find the distance between the towers. Solution:

      A               B
      |30m            |50m
      | \ θ       φ  /|
      |  \           / |
      |   \         /  |
      |    \       /   |
      +-----P-----+    
          d   d

Since angles are complementary:

$$\tan\theta\tan\phi=1$$ $$\frac{30}{d}\times\frac{50}{d}=1$$ $$d^2=1500$$ $$d=10\sqrt{15}$$

Total distance:

$$2d=\boxed{20\sqrt{15}\text{ m}}$$

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7. Moving Observer Tracking

• Formula:

$$h=d_1\tan\theta_1=d_2\tan\theta_2$$

If the observer moves (x) m towards the object:

$$d_2=d_1-x$$

• Example: From point A, the angle of elevation of a tower is (30^\circ). After moving 30 m towards it, the angle becomes (45^\circ). Find the height. Solution:

          T
          |\
        h | \
          |  \
          |   \
          B----A------C
             d    30m

Let the original distance be (d).

From A:

$$h=d\tan30^\circ=\frac d{\sqrt3}$$

After moving 30 m:

$$h=(d-30)\tan45^\circ=d-30$$

Therefore:

$$\frac d{\sqrt3}=d-30$$ $$d(\sqrt3-1)=30\sqrt3$$ $$d=\frac{30\sqrt3}{\sqrt3-1} =15(3+\sqrt3)$$

Hence:

$$h=\frac d{\sqrt3} =\boxed{15(1+\sqrt3)\text{ m}}$$

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8. Equivalent Single Discount

• Formula: For successive discounts (d_1%) and (d_2%):

$$D_{\text{eq}} =d_1+d_2-\frac{d_1d_2}{100}$$

More generally:

$$SP=MP\left(1-\frac{d_1}{100}\right) \left(1-\frac{d_2}{100}\right)$$

• Example: Find the single equivalent discount of 20% and 10%. Solution:

$$D_{\text{eq}} =20+10-\frac{20\times10}{100}$$ =30-2 =\boxed{28%}

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9. Three or More Successive Discounts

• Formula:

$$SP=MP\prod_{i=1}^{n}\left(1-\frac{d_i}{100}\right)$$

Equivalent discount:

$$D_{\text{eq}} =\left[1-\prod_{i=1}^{n} \left(1-\frac{d_i}{100}\right)\right]\times100$$

• Example: Find the equivalent discount of 10%, 20% and 25%. Solution:

$$SP=MP(0.9)(0.8)(0.75)$$ $$=0.54MP$$

Thus customer pays 54% of MP.

D_{\text{eq}}=100-54 =\boxed{46%}

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10. Discount vs. Profit Margin Linkage

• Formula:

$$SP=CP\left(1+\frac{p}{100}\right)$$

and:

$$SP=MP\left(1-\frac d{100}\right)$$

Therefore:

$$MP\left(1-\frac d{100}\right) =CP\left(1+\frac p{100}\right)$$

• Example: A shopkeeper marks an article 40% above CP and gives a 20% discount. Find the profit percentage. Solution:

Let:

$$CP=100$$

Then:

$$MP=140$$

After 20% discount:

$$SP=140(0.8)=112$$

Profit:

$$112-100=12$$ \boxed{\text{Profit}=12%}

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11. Marked Price from Cost Price and Desired Profit

• Formula: If desired profit is (p%) and discount is (d%):

$$MP= \frac{CP(1+p/100)} {1-d/100}$$

• Example: An article costs ₹800. A seller wants a 20% profit after giving a 20% discount. Find the required marked price. Solution:

Required SP:

$$800(1.2)=₹960$$

Since 20% discount means:

$$SP=0.8MP$$

Therefore:

$$MP=\frac{960}{0.8} =\boxed{₹1200}$$

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12. Reverse Discount Calculations

• Formula: For successive discounts:

$$MP= \frac{SP} {(1-d_1/100)(1-d_2/100)\cdots}$$

• Example: After successive discounts of 10% and 20%, the selling price is ₹720. Find the marked price. Solution:

$$720=MP(0.9)(0.8)$$ $$720=0.72MP$$ $$MP=\frac{720}{0.72} =\boxed{₹1000}$$

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13. Finding Discount from MP and SP

• Formula:

$$d=\frac{MP-SP}{MP}\times100$$

• Example: An article marked at ₹2,500 is sold for ₹2,000. Find the discount percentage. Solution:

Discount:

$$2500-2000=₹500$$ d=\frac{500}{2500}\times100 =\boxed{20%}

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14. Finding Profit/Loss After Discount

• Formula:

\text{Profit%} =\frac{SP-CP}{CP}\times100 \text{Loss%} =\frac{CP-SP}{CP}\times100

• Example: An article costs ₹1,000, is marked at ₹1,500 and sold at a 20% discount. Find the profit percentage. Solution:

$$SP=1500(0.8)=₹1200$$

Profit:

$$1200-1000=₹200$$ \text{Profit%} =\frac{200}{1000}\times100 =\boxed{20%}

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Advanced Variants

15. Required Markup for a Given Profit After Discount

• Formula: If markup is (m%), discount is (d%), and desired profit is (p%):

$$(1+m/100)(1-d/100)=1+p/100$$

Hence:

$$m= \left[ \frac{1+p/100}{1-d/100}-1 \right]\times100$$

• Example: What markup should a seller use to earn 20% profit after giving a 25% discount? Solution:

$$(1+m/100)(0.75)=1.20$$ $$1+\frac m{100}=\frac{1.20}{0.75}=1.6$$ $$m=60$$ \boxed{60%}

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16. Discount Equivalent to a Loss

• Formula: If an article is marked (m%) above CP and sold at discount (d%):

$$\frac{SP}{CP} =\left(1+\frac m{100}\right) \left(1-\frac d{100}\right)$$

If this ratio is less than 1, there is a loss.

• Example: An article is marked 20% above CP and sold at a 25% discount. Find the loss percentage. Solution:

Let:

$$CP=100$$ $$MP=120$$ $$SP=120(0.75)=90$$

Loss:

$$100-90=10$$ \boxed{\text{Loss}=10%}

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17. Two Successive Discounts vs. One Discount

• Formula:

$$d_{\text{eq}} =d_1+d_2-\frac{d_1d_2}{100}$$

The equivalent single discount always gives the same final SP as the successive discounts.

• Example: Is a single 30% discount equivalent to successive discounts of 20% and 10%? Solution:

$$D_{\text{eq}} =20+10-\frac{20\times10}{100}$$ $$=28$$

Therefore:

\boxed{\text{No; successive discounts are equivalent to }28%}

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18. Profit Percentage When Discount Is Given as a Fraction of MP

• Formula: If discount is (d%):

$$SP=MP\left(1-\frac d{100}\right)$$

Then compare SP with CP.

• Example: An article is marked 50% above CP and sold at a 10% discount. Find the profit percentage. Solution:

Let:

$$CP=100$$ $$MP=150$$ $$SP=150(0.9)=135$$

Profit:

$$135-100=35$$ \boxed{\text{Profit}=35%}