We read your photo as:

To construct triangle PQR and measure the lengths of PQ and PR:

Mathematics

ScanSolve AI Answer

Answer

∣PQ∣=3.11 cm|PQ| = \text{3.11 cm}

∣PR∣=6.00 cm|PR| = \text{6.00 cm}

Step-by-step solution

4 steps

To construct triangle PQR and measure the lengths of PQPQ and PRPR:

Given:

  • QR=6 cmQR = 6 \text{ cm}
  • ∠PQR=75∘\angle PQR = 75^\circ
  • ∠PRQ=30∘\angle PRQ = 30^\circ

Construction Steps:

  1. Draw the base QRQR.

    Draw a straight line and mark a point Q on it. Using a ruler, measure 6 cm6 \text{ cm} from Q and mark point R. So, QR=6 cmQR = 6 \text{ cm}.

  2. Construct ∠PRQ=30∘\angle PRQ = 30^\circ at point R.

    • With R as the center, draw an arc of any convenient radius to intersect QRQR at a point, say A.
    • With A as the center and the same radius, draw another arc to intersect the first arc at a point, say B. ∠BRQ=60∘\angle BRQ = 60^\circ.
    • Bisect ∠BRQ\angle BRQ. With A and B as centers, draw arcs of the same radius (larger than half the distance AB) to intersect at a point, say C.
    • Draw a line from R through C. This line forms an angle of 30∘30^\circ with QRQR.
  3. Construct ∠PQR=75∘\angle PQR = 75^\circ at point Q.

    • With Q as the center, draw an arc of any convenient radius to intersect QRQR at a point, say D.
    • With D as the center and the same radius, draw an arc to intersect the first arc at E (60∘60^\circ).
    • With E as the center and the same radius, draw an arc to intersect the first arc at F (120∘120^\circ).
    • Bisect the angle between E and F to construct 90∘90^\circ. With E and F as centers, draw arcs of the same radius to intersect at G.
    • Draw a dashed line from Q through G. This line represents 90∘90^\circ. Let the intersection of this line with the initial arc be H.
    • Now, bisect the angle between E (60∘60^\circ) and H (90∘90^\circ). With E and H as centers, draw arcs of the same radius to intersect at I.
    • Draw a line from Q through I. This line forms an angle of 60∘+15∘=75∘60^\circ + 15^\circ = 75^\circ with QRQR.
  4. Locate point P.

    The intersection of the line from R (from Step 2) and the line from Q (from Step 3) is point P. Connect P to Q and P to R to complete triangle PQR.

    Measurement of ∣PQ∣|PQ| and ∣PR∣|PR|:

    Since physical measurement cannot be performed by an AI, we will calculate the lengths using trigonometry.

    First, find the third angle ∠QPR\angle QPR:
    The sum of angles in a triangle is 180∘180^\circ.

    ∠QPR=180∘−∠PQR−∠PRQ\angle QPR = 180^\circ - \angle PQR - \angle PRQ ∠QPR=180∘−75∘−30∘\angle QPR = 180^\circ - 75^\circ - 30^\circ ∠QPR=180∘−105∘\angle QPR = 180^\circ - 105^\circ ∠QPR=75∘\angle QPR = 75^\circ

    Since ∠PQR=∠QPR=75∘\angle PQR = \angle QPR = 75^\circ, triangle PQR is an isosceles triangle with sides opposite these angles being equal.
    Therefore, PR=QRPR = QR.
    Given QR=6 cmQR = 6 \text{ cm},

    PR=6 cmPR = 6 \text{ cm}

    Now, use the Sine Rule to find the length of PQPQ:

    PQsin⁡(∠PRQ)=QRsin⁡(∠QPR)\frac{PQ}{\sin(\angle PRQ)} = \frac{QR}{\sin(\angle QPR)} PQsin⁡(30∘)=6 cmsin⁡(75∘)\frac{PQ}{\sin(30^\circ)} = \frac{6 \text{ cm}}{\sin(75^\circ)} PQ=6 cm×sin⁡(30∘)sin⁡(75∘)PQ = \frac{6 \text{ cm} \times \sin(30^\circ)}{\sin(75^\circ)}

    We know sin⁡(30∘)=0.5\sin(30^\circ) = 0.5 and sin⁡(75∘)=6+24≈0.9659\sin(75^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659.

    PQ=6 cm×0.50.9659PQ = \frac{6 \text{ cm} \times 0.5}{0.9659} PQ=3 cm0.9659PQ = \frac{3 \text{ cm}}{0.9659} PQ≈3.1058 cmPQ \approx 3.1058 \text{ cm}

    Rounding to two decimal places:

    PQ≈3.11 cmPQ \approx 3.11 \text{ cm}

    The measured lengths are:
    ∣PQ∣=3.11 cm|PQ| = \boxed{\text{3.11 cm}}
    ∣PR∣=6.00 cm|PR| = \boxed{\text{6.00 cm}}

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp