Question

Refer the attachment and answer the questions.
Four questions are given,
Solve any 3 of them.

#️⃣ No spamming #️⃣​

Answer 1

Hey mate,

Refer to the above attachment ⤴ for your reference,

i) Construction :

Draw a perpendicular from vertex T to the extended line RK at Z,

So according to the figure TZ becomes the common height for triangles TRP and TPK.

So we have,

RP:PK = 3:2

Now removing ratio,

RP = 3x,

PK = 2x

Now we know according to the property,

Ratio of areas of two triangles with common height = ratio of corresponding bases,

Triangle(TRK) : Triangle (TPK) = RP:PK

= 3:2

iii) Angle given = - 60°

So cosx = cos(-60°)

We know,

Cos(-x) = cosx

So,

= 1/2

iv) We know,

Slope = tanx

So,

Tan(45°) = 1

Hope this helps you out!

Answer 2

Answer:

i. $\boxed{\red{\sf\:A\:(\:\triangle\:TRP\:)\::\:A\:(\:\triangle\:TPK\:)\:=\:3\::\:2}}$

ii. $\boxed{\pink{\sf\:d\:(\:C_{1}\:,\:C_{2}\:)\:=\:11\:units}}$

iii. $\boxed{\blue{\sf\:cos\:\theta\:=\:\frac{1}{2}}}$

iv. $\boxed{\orange{\sf\:Slope\:of\:line\:=\:1}}$

Step-by-step-explanation:

i.

$\sf\:\triangle\:TRP$ and $\sf\:\triangle\:TPK$ are acute angled triangles.

We know that, heights of the acute angled triangles are inside the triangle.

$\sf\therefore\:Both\:triangles\:have\:equal\:heights$

Ratio of areas of two triangles with equal heights is equal to the ratio of their corresponding bases. [ Property ]

$\sf\therefore\:\dfrac{A\:(\:\triangle\:TRP\:)}{A\:(\:\triangle\:TPK\:)}\:=\:\dfrac{RP}{PK}\:=\:\dfrac{3}{2}\\\\\implies\sf\:\dfrac{A\:(\:\triangle\:TRP\:)}{A\:(\:\triangle\:TPK\:)}\:=\:\frac{3}{2}\\\\\boxed{\red{\sf\:A\:(\:\triangle\:TRP\:)\::\:A\:(\:\triangle\:TPK\:)\:=\:3\::\:2}}$

__________________________

ii.

Two circles of radii 8 units and 3 units touch each other externally.

We know that,

The distance between the centres of two circles touching externally is equal to the sum of their radii.

$\sf\therefore\:d\:(\:C_{1}\:,\:C_{2}\:)\:=\:r_{1}\:+\:r_{2}\\\\\implies\sf\:d\:(\:C_{1}\:,\:C_{2}\:)\:=\:8\:+\:3\:=\:11\:units\\\\\boxed{\pink{\sf\:d\:(\:C_{1}\:,\:C_{2}\:)\:=\:11\:units}}$

_____________________

iii.

$\sf\:\theta\:=\:-\:60^{\circ}$

We know that,

$\sf\:cos\:(\:-\:\alpha\:)\:=\:cos\:\alpha\\\\\implies\sf\:cos\:(\:-\:60^{\circ}\:)\:=\:cos\:60^{\circ}\\\\\implies\sf\:cos\:(\:-\:60^{\circ}\:)\:=\:\frac{1}{2}\:\:\:-\:-\:-\:[\:From\:trigonometric\:table\:]\\\\\implies\sf\:cos\:(\:-\:60^{\circ}\:)\:=\:cos\:\theta\:=\:\frac{1}{2}\\\\\boxed{\blue{\sf\:cos\:\theta\:=\:\frac{1}{2}}}$

____________________

iv.

We have given that, the inclination of line is 45°.

It means, $\sf\:\theta\:=\:45^{\circ}$

We know that,

$\sf\:Slope\:of\:line\:=\:tan\:\theta\\\\\implies\sf\:Slope\:of\:line\:=\:tan\:45^{\circ}\\\\\boxed{\orange{\sf\:Slope\:of\:line\:=\:1}}\:\:\:-\:-\:-\:[\sf\:From\:trigonometric\:table\:]$