Let PQRS be a quadrilateral in a plane, where QR = 1, $\angle PQR=\angle QRS=70°$, $\angle PQS=15°$ and $\angle PRS=40°$. If $\angle RPS=\theta °$, $PQ=\alpha$ and $PS=\beta$, then the interval(s) that contain(s) the value of $4\alpha \beta sin\theta °$ is/are

The correct options are (0, √2) and (1, 2).

We are given quadrilateral PQRS with QR = 1, ∠PQR = ∠QRS = 70°, ∠PQS = 15°, and ∠PRS = 40°. Let ∠RPS = θ°, PQ = α, PS = β. Our goal is to find the exact value of 4αβ sin θ°.

Step 1: Identify sub-angles using the diagonals PR and QS.

Since QS is a diagonal, it splits the angle at Q:

∠SQR = ∠PQR - ∠PQS = 70° - 15° = 55°

Since PR is a diagonal, it splits the angle at R:

∠PRQ = ∠QRS - ∠PRS = 70° - 40° = 30°

Step 2: Analyze triangle PQR to find α = PQ.

In △PQR, the three angles are:

∠PQR = 70°,   ∠PRQ = 30°,   ∠QPR = 180° - 70° - 30° = 80°

Applying the Sine Rule with QR = 1:

\frac{\mathrm{PQ}}{sin(30°)}=\frac{1}{sin(80°)}

\alpha =\frac{sin(30°)}{sin(80°)}=\frac{1}{2sin(80°)}

Step 3: Analyze triangle QRS to find RS.

In △QRS:

∠SQR = 55°,   ∠QRS = 70°,   ∠QSR = 180° - 55° - 70° = 55°

Since ∠SQR = ∠QSR = 55°, triangle QRS is isosceles, giving us:

RS = QR = 1

Step 4: Analyze triangle PRS to relate β and θ.

In △PRS, we have RS = 1, ∠PRS = 40°, ∠RPS = θ°. Applying the Sine Rule:

\frac{\mathrm{PS}}{sin(40°)}=\frac{\mathrm{RS}}{sin(\theta °)}=\frac{1}{sin(\theta °)}

This gives us the critical relation:

\beta ·sin(\theta °)=sin(40°)

Step 5: Compute 4αβ sin θ°.

Substituting the relation from Step 4:

4\alpha \beta sin\theta °=4\alpha ·sin(40°)

Substituting α from Step 2:

=4·\frac{1}{2sin(80°)}·sin(40°)=\frac{2sin(40°)}{sin(80°)}

Now apply the double-angle identity: sin(80°) = 2 sin(40°) cos(40°):

=\frac{2sin(40°)}{2sin(40°)cos(40°)}=\frac{1}{cos(40°)}=sec(40°)

Step 6: Evaluate numerically and identify the interval(s).

cos(40°) ≈ 0.7660, so:

4\alpha \beta sin\theta °=\frac{1}{cos(40°)}\approx 1.3054

Now check each interval:

• (0, √2): Since √2 ≈ 1.4142, and 0 < 1.3054 < 1.4142 ✓  Contains the value.
• (1, 2): Since 1 < 1.3054 < 2 ✓  Contains the value.
• (√2, 3): Since 1.3054 < √2 ≈ 1.4142, 1.3054 ∉ (√2, 3) ✗
• (2√2, 3√2): Since 2√2 ≈ 2.828, clearly 1.3054 < 2.828 ✗

Therefore, the value of 4\alpha \beta sin\theta °=sec(40°)\approx 1.305 is contained in the intervals (0, √2) and (1, 2).