The rate at which the area of the triangle is increasing when the angle between the sides of fixed lengths is π/3 is 0.3 m²/s.
What is the rate of change of an area?
If the lengths a and b for two sides of the triangle, as well as the included angle θ, are known, the area of a triangle can be calculated using the sine formula.
[tex]\text { Area }=\frac{1}{2} a b \sin \theta[/tex]
Now, according to the question;
The two sides of the triangle are given;
Let a = 4m and b = 5m.
Let θ = π/3. be the angle between the two sides;
Then,
Differentiate the area with respect to time (as rate of change of area)
[tex]\begin{aligned}\frac{\mathrm{d} A}{\mathrm{~d} t}=\frac{\mathrm{d}}{\mathrm{d} t} \frac{1}{2} a b \sin \theta &=\frac{1}{2} * 4 * 5 * \frac{\mathrm{d}(\sin \theta)}{\mathrm{d} \theta} \cdot \frac{\mathrm{d} \theta}{\mathrm{d} t} \\&=10 *(\cos \theta) * 0.06 \frac{\mathrm{m}^{2}}{\mathrm{~s}}=0.6 \cos (\pi / 3) \frac{\mathrm{m}^{2}}{\mathrm{~s}} \\&=0.6 * 0.5 \frac{\mathrm{m}^{2}}{\mathrm{~s}}=0.3 \frac{\mathrm{m}^{2}}{\mathrm{~s}}\end{aligned}[/tex]
Therefore, the area of the triangle as a function of angle is 0.3 m²/s.
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The complete question is-
Two sides of a triangle are 4m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed lengths is π/3.
