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Ask QuestionPosted by Aman Kumar 6 years, 3 months ago
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Posted by ? ? 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
Let height of flagstaff be BD = x m
{tex}\therefore \quad \tan 30 ^ { \circ } = \frac { A B } { A P }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { 10 } { A P }{/tex}
{tex}A P = 10 \sqrt { 3 }{/tex}
distance of the building from P.
{tex}= 10 \times 1.732 = 17.32 \mathrm { m }{/tex}
{tex}\tan 45 ^ { \circ } = \frac { A D } { A P } \text { or } 1 = \frac { 10 + x } { 17 \cdot 32 }{/tex}
or, x + 10 = 17.32
x = 17.32 - 10
x = 7.32
Hence, length of flagstaff x = 1 7.32 m.
Posted by Adarsh Roy 6 years, 3 months ago
- 1 answers
Sia ? 6 years, 3 months ago
{tex}\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = \frac{{\tan A + \sec A - ({{\sec }^2}A - {{\tan }^2}A)}}{{\tan A - \sec A + 1}}{/tex} [{tex}\because{/tex} 1 + tan2A = sec2A {tex}\Rightarrow{/tex} sec2A - tan2A = 1]
{tex} = \frac{{\tan A + \sec A - \{ (\sec A - \tan A)(\sec A + \tan A)\} }}{{\tan A - \sec A + 1}}{/tex}
Now, take (tanA + secA) as a common term, we get
{tex} = \frac{{(\tan A + \sec A)(1 - \sec A + \tan A)}}{{\tan A - \sec A + 1}}{/tex}
= tanA + secA
{tex} = \frac{{\sin A}}{{\cos A}} + \frac{1}{{\cos A}}{/tex}
{tex} = \frac{{1 + \sin A}}{{\cos A}}{/tex}
{tex}\therefore{/tex} Hence proved
Posted by Lakshay Kochhar 6 years, 3 months ago
- 3 answers
Yogita Ingle 6 years, 3 months ago
Let us assume that 5+2√3 is rational
5+2√3 = p/q ( where p and q are co prime)
2√3 = p/q-5
2√3 = p-5q/q
√3 = p-5q/2q
now p , 5 , 2 and q are integers
∴ p-5q/2q is rational
∴ √3 is rational
but we know that √3 is irrational . This is a contradiction which has arisen due to our wrong assumption.
∴ 5+2√3 is irrational
Ravenclaw_? Raunak 6 years, 3 months ago
Posted by Iramnaaz Ansari 6 years, 3 months ago
- 1 answers
Posted by Aarushi Singhal 6 years, 3 months ago
- 1 answers
Sia ? 6 years, 3 months ago
Let us assume, to the contrary, that is {tex}3 + \sqrt { 5 }{/tex} rational.
That is, we can find coprime integers a and b {tex}( b \neq 0 ){/tex} such that
{tex}3 + \sqrt { 5 } = \frac { a } { b } \text { Therefore, } \frac { a } { b } - 3 = \sqrt { 5 }{/tex}
{tex}\Rightarrow \frac { a - 3 b } { b } = \sqrt { 5 }{/tex}
{tex}\Rightarrow \frac { a - 3 b } { b } = \sqrt { 5 } \Rightarrow \frac { a } { b } - \frac { 3 } { 2 }{/tex}
Since a and b are integers,
We get {tex}\frac { a } { b } - \frac { 3 } { 2 }{/tex} is rational, also so {tex}\sqrt { 5 }{/tex} is rational.
But this contradicts the fact that {tex}\sqrt { 5 }{/tex} is irrational.
This contradiction arose because of our incorrect
assumption that {tex}3 + \sqrt { 5 }{/tex} is rational.
So, we conclude that {tex}3 + \sqrt { 5 }{/tex} is irrational.
Posted by Abc Def 6 years, 3 months ago
- 2 answers
Gaurav Seth 6 years, 3 months ago
Given that the angle of depression is 30° , then angle of elevation is (90 - 30 = 60°). Therefore,
We have,
tan60° = height of tower /distance of the car from the base of the tower
=>√3 = 75/distance of the car from the base of the tower
=>distance of the car from the base of the tower = 75/ √3 or 75√3/(√3√3) = 25√3
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Aman Kumar 6 years, 3 months ago
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