In - RST in the figure- what is the value of x-

In Δ RST OR and OT are the bisectors of ∠ R and ∠ T respectively ∴∠ ROT=90^∘+1/2∠ S ⇒ 140^∘=90^∘+1/2x ⇒1/2x=140^∘ 90^∘=50^∘ x=50^∘× 2=100^∘ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Triangle and Sum of Its Internal Angles

In Δ RST in t...

Question

In $Δ R S T$ (in the figure), what is the value of x?

Open in App

Solution

In $Δ R S T$

OR and OT are the bisectors of $∠ R$ and $∠ T$ respectively

$∴ ∠ R O T = 90^{\circ} + \frac{1}{2} ∠ S$

$\Rightarrow 140^{\circ} = 90^{\circ} + \frac{1}{2} x$

$\Rightarrow \frac{1}{2} x = 140^{\circ} - 90^{\circ} = 50^{\circ}$

$x = 50^{\circ} \times 2 = 100^{\circ}$

Suggest Corrections

Similar questions

Q. In ΔRST (See figure), what is the value of x?

(a) 40

(b) 90°

(c) 80°

(d) 100

Q. In the figure above, PQRS is a rectangle. The area of

Δ R S T

7

and

P T = \frac{2}{5} P S

. What is the area of

□ P Q R S

Q. In figure, altitude RH = 15 and

¯ ¯¯¯¯¯ ¯ S T

is drawn parallel to

¯ ¯¯¯¯¯¯ ¯ Q P

, What must be the length of

¯ ¯¯¯¯¯¯ ¯ R J

so that the area of

Δ R S T = \frac{1}{3}

the area of

Δ

RQP ?

$△ R S T \sim △ X Y Z .$ In $△ R S T$ , $R S = 4.5$ cm, $∠ R S T = 40^{\circ}$ , $S T = 5.7$ cm Construct $△ R S T$ and $△ X Y Z$ , such that $\frac{R S}{X Y} = \frac{3}{5}$ .

In the figure, what is the value of x?

Join BYJU'S Learning Program

In ΔRST (in the figure), what is the value of x?

In ΔRST OR and OT are the bisectors of ∠R and ∠T respectively ∴∠ROT=90∘+12∠S ⇒140∘=90∘+12x ⇒12x=140∘−90∘=50∘ x=50∘×2=100∘

In $Δ R S T$ (in the figure), what is the value of x?

In $Δ R S T$

OR and OT are the bisectors of $∠ R$ and $∠ T$ respectively

$∴ ∠ R O T = 90^{\circ} + \frac{1}{2} ∠ S$

$\Rightarrow 140^{\circ} = 90^{\circ} + \frac{1}{2} x$

$\Rightarrow \frac{1}{2} x = 140^{\circ} - 90^{\circ} = 50^{\circ}$

$x = 50^{\circ} \times 2 = 100^{\circ}$