Math, asked by ashaaroraa2, 1 month ago

In a triangle PQR if PQ=PR and ∠QPR=30° Then find the measure of ∠PQR.


full explanations please ​

Answers

Answered by knand3789
1

Answer:

Solution

Correct option is

B

The area of the triangle PQR is 25

3

and ∠QRP=120

o

C

The radius of the incircle of the triangle PQR is 10

3

−15

D

The area of the circumcircle of the triangle PQR is 100π

(A) cos30

o

=

2×10

3

×10

(10

3

)

2

+(10)

2

−(PR)

2

2

3

=

200

3

400−(PR)

2

⇒PR=10

∵QR=PR ⇒PQR=∠QPR

∠QPR =30

o

(B) Area of △PQR =

2

1

×10

3

×10×sin30

o

=

2

1

×10×10

3

×

2

1

=25

3

∠QRP=180

o

−(30

o

+30

o

)=120

o

(C) r=

S

=

(

2

10+10+10

3

)

25

3

=

10+5

3

25

3

=5

3

⋅(2−

3

)=10

3

−15

(D) R=

2sinA

a

=

2sin30

o

10

=10

∴ Area =πR

2

=100π

Step-by-step explanation:

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Answered by sangram0111
1

Given:

In a triangle PQR if PQ=PR and ∠QPR=30°

Solution:

Know that angles opposite to the equal sides of a triangle are also equal.

Refer the figure given below,

Since, \[PQ = PR\]

\[\therefore \angle PQR = \angle PRQ\]

In \[\Delta PQR\],

\[ \Rightarrow \angle PQR + \angle PRQ + \angle QPR = 180^\circ \]

Put,\[\angle QPR = 30^\circ \] and \[\angle PRQ = \angle PQR\]

\[ \Rightarrow 2 \times \angle PQR + 30^\circ  = 180^\circ \]

\[ \Rightarrow \angle PQR = \frac{{150^\circ }}{2}\]

\[ \Rightarrow \angle PQR = 75^\circ \]

Hence, the measure of angle PQR is \[75^\circ \].

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