Given: △PQR, m∠R = 90°, MP = 18 m∠PQR = 75°, m∠MQR = 60° Find: RQ

Question

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Respuesta :

ColinJacobus ColinJacobus · Answer 1 · 2018-11-20T11:53:05+01:00

Correct answer is RQ=9

Solution:-

Given that in a ΔPQR, m∠R=90°, MP=18, m∠PQR=75° and m∠MQR=60°.

Since m∠MQR<m∠PQR, M should lie on PR only, Please refer the diagram attached for reference.

From the diagram∠PQM=∠PQR-∠MQR=75°-16°=15°

And in ΔPQR,∠RPQ=180-(remaining 2 angles in triangle)

=180-(90+75)=15°

So, in ΔPQM, ∠MPQ=∠PQM hence the opposite sides must be equal that is

MP=MQ but given MP=18

Hence MQ=18

In ΔMRQ, cos(∠MQR)=[tex]\frac{RQ}{MQ}[/tex]

cos(60°)=[tex]\frac{RQ}{18}[/tex]

RQ= 18cos(60°)=18X0.5 =9

Hence RQ=9

apocritia apocritia · Answer 2 · 2019-01-06T04:05:06+01:00

Answer:

RQ = 9

Step-by-step explanation:

In the given triangle,

∠R = 90° , ∠PQR = 75° , ∠MQR = 60° and MP = 18

From the diagram

∠RPQ = ∠ PRQ - ∠PQR = 90° - 75° = 15°·············(1)

In ΔMPQ

∠MQP = ∠PQR - ∠MQR = 75° - 60° = 15°·················(2)

from (1) and (2)

∠RPQ = ∠MQP = 15°

So, ΔPMQ is an isosceles triangle.

Hence,

MP = MQ = 18

Now, in ΔMRQ

cos(60°) = [tex]\frac{RQ}{MQ}[/tex]

[tex]\frac{1}{2}[/tex] = [tex]\frac{RQ}{18}[/tex]

RQ = 0.5 × 18

RQ = 9