Height and Distance formulas with examples for SSC and Bank Exams

Vikram Singh5 years ago 5.3K Views Join Examsbookapp store google play
height and distance formulas

If you are finding hard to use height and distance formulas in ssc or bank exams, then no need to worry, I am sharing height and distance formulas with examples which will help you in your exam preparations.

Here in this blog, you can learn easily how to use height and distance formulas and know how many types of formulas you can use with different-2 height and distance equational examples. Click here to more practice of height and distance with questions and answers.


Height and Distance Formulas with Examples for Competitive Exams


1. We already know that :

In a right triangle . ∆ OAB, where  ∠ BOA = θ, 

$$ A. \ sinθ = {Prependicuar\over Hypotenuse}={AB\over OB}; $$

   $$ B. \ cosθ = {Base\over Hypotenuse}={OA\over OB}; $$

   $$ C. \ tanθ = {Prependicuar\over Base}={AB\over OA}; $$

   $$ D. \ cosecθ = {1\over sinθ}={OB\over AB}; $$

    $$ E. \ secθ = {1\over cosθ}={OB\over OA}; $$

  $$ F. \ cotθ = {1\over tanθ}={OA\over AB}; $$

2. Trigonometrical Identities :

(1) sin2θ + cos2θ = 1

(2) 1 + tan2θ = sec2θ

(3) 1 + cot2θ = cosec2θ

3. Values of T-ratios

4. Angle of Elevation : Suppose a man from a point O looks up at an object P. placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.

∴ Angle of elevation P from O =  ∠AOP.


5. Angle of Depression : Suppose a man from a point O looks down at an object P, placed below the level of his eye, then the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

Q.1.The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree, is : 

(A) 300        (B) 450        (C) 600                   (D)900

Explanation 


Let AB be the tree and AC be its shadow.

Let ∠ACB = θ

  $$ Then,{AC\over AB}= \sqrt {3 } \ ⇒ θ = 30^0  $$

Q.2. From a point P on a level ground, the angle of elevation of the top of a tower is 300. If the tower is 100 m high, the distance of point P from the foot of the tower is: 

(A) 149 m             (B) 156 m             (C) 173 m             (D) 200 m

Explanation 


Let AB be the tower. Then, ∠APB = 300 and AB = 100 m

  $$ {AB\over AP}= tan30^0 =\ {1\over \sqrt { 3} \ }  \ ⇒ AP = (AB×\sqrt {3 } \ )=100\sqrt { 3} \   $$

  $$=(100×1.73) \ m =173 \ m. $$

Q.3 The angle of elevation of a ladder leaning against a wall is 600 and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is : 

(A) 2.3 m              (B) 4.6 m              (C) 7.8 m              (D) 9.2 m 

Explanation 


Let AB be the well and BC be the ladder.

Then, ∠ACB = 600 and AC = 4.6 m.

$$  {AC\over BC}=cos60^0={1\over2}$$

⇒ BC = 2×AC = (2×4.6) m = 9.2 m.

Q.4.An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 300. The height of the tower is: 

(A) 21.6 m         (B) 23.2 m         (C) 24.72 m       (D) None of these 

Explanation


Let AB be the observer and CD be the tower Draw BE⊥CD.

Then, CE = AB = 1.6 m, BE = AC = 2√3 m.

  $$  {DE\over BE}=tan \ 30^0={1\over \sqrt{3} \ }⇒ DE={20\sqrt{3} \ \over \sqrt{3} \ }m=20 \ m $$

∴ CD = CE+DE = (1.6+20)m = 21.6 m.

Q.5.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the two ships are 300 and 450 respectively. If the lighthouse is 100 m high, the distance between the two ships is : 

(A) 173 m          (B) 200 m          (C) 273 m           (D) 300 m 

Explanation


Let AB be the lighthouse and C and D be the positions of the ships. Then,

AB = 100 m, ∠ACB = 300 and ∠ADB = 45

$$ {AB\over AC}=tan \ 30^0={1\over \sqrt{3} \ } $$

  $$⇒ AC = AB× \sqrt {3 } \ =100\sqrt { 3} \ m$$

$$ {AB\over AD}=tan \ 45^0 =1⇒AD=AB=100 \ m$$

$$ ∴ CD = (AC+AD)=(100\sqrt { 3} \ +100 ) \ m $$

  $$ =100(\sqrt { 3}+1) \ m=(100×273)m=273 $$

I hope these formulas will helpful for your score high. You can ask me anything in the comment section related height and distance formulas with examples. Visit next page for more Height and Distance Formulas with Examples

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    Vikram Singh

    Providing knowledgable questions of Reasoning and Aptitude for the competitive exams.

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