Question

answer my five questions
I will thank your all answers
I will follow u
make your answers as brainliest
but you put irrelevant answers
you will be reported ​

Answer 1

1

 The distance between two points is 5 units.

$\sf{\implies \sqrt{(5-1)^2+(-2-a)^2}=5}$

$\sf{\implies \sqrt{16+(-2-a)^2}=5}$

$\sf{\implies 16+(-2-a)^2=25}$

$\sf{\implies (-2-a)^2=9}$

$\sf{\implies -2-a=\pm 3}$

$\sf{\implies a=1,-5}$

2

 If it is colinear, point B should lie on $\sf{\overline{AC}}$. Let point B divides

$\sf{\implies B(\dfrac{13k+7}{k+1} ,\dfrac{k-5}{k+1} )}$

$\sf{\implies \dfrac{13k+7}{k+1} =9,\dfrac{k-5}{k+1} =-3}$

$\sf{\implies 13k+7=9k+9, k-5=-3k-3}$

$\sf{\implies 4k-2=0, 4k-2=0}$

 Both equations have equal solutions. So point B lies on $\sf{\overline{AC}}$.

3

 The elevation is the ratio of height and base. It is related to tangent value.

 The ratio of the two sides is $\sf{1:\sqrt{3} }$.

 Therefore, the elevation is $\sf{\dfrac{1}{\sqrt{3} } }$.

⇒ If the angle of elevation is θ, the elevation is tan θ.

$\sf{\implies tan \theta=\dfrac{1}{\sqrt{3} } }$

 The angle of elevation is θ=30°.

4

 The midpoint of two points can be found.

$\implies \displaystyle{(\frac{-2-6}{2} ,\frac{3-5}{2} )}$

$\implies \displaystyle{(-4,-1)}$

5

 $\sf{cos\:\theta=\dfrac{base}{hypotenuse} }$

 The ratio of base and hypotenuse is 3:4.

 Let given sides be 3k and 4k.

 The height is √7k.

 The ratio of the three sides is √7:3:4.

• $\sf{sin\:A=\dfrac{height}{hypotenuse} =\dfrac{\sqrt{7} }{4} }$
• $\sf{cos\:A=\dfrac{base}{hypotenuse} =\dfrac{3}{4} }$
• $\sf{tan\:A=\dfrac{height}{base} =\dfrac{\sqrt{7} }{3} }$

Given: $\sf{\dfrac{sin\:A-cos\:A}{2tan\:A} }$

$\sf{=(\dfrac{\sqrt{7} }{4} -\dfrac{3}{4} )\times \dfrac{3}{2\sqrt{7} } }$

$\sf{=\dfrac{\sqrt{7} -3}{4} \times \dfrac{3}{2\sqrt{7} } }$

$\sf{=\dfrac{3\sqrt{7} -9}{8\sqrt{7} } }$

$\sf{=\dfrac{\sqrt{7} \times (3\sqrt{7} -9)}{8\sqrt{7} \times \sqrt{7} } }$

$\sf{=\dfrac{21 -9\sqrt{7} }{56} }$