Question

A triangle has two sides a = 1cm and b=2 cm. How fast is the third
side c increasing when the angle a between the given sides is 60°
and is increasing at the rate of 3° per second ? [Hint : Use cosine
rule]
lit
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Answer 1

The third side c is increasing at $3$ cm/sec when the angle between the given sides is $60$° and is increasing at the rate of $3$° per second

Step-by-step explanation:

Given: A triangle has two sides a = $1$ cm and b=$2$ cm and the angle between the given sides is $60$° and is increasing at the rate of $3$° per second

To Find: How fast is the third side c increasing when the angle a between the given sides is $60$° and is increasing at the rate of $3$° per second ?

Concept/Formula used: By Cosine Rule

Considering Δ ABC

By using Cosine Rule

$c^2=a^2+b^2-2abcosc$

$c=\sqrt{3}$ cm when ∠c$=60$°

$c^2=a^2+b^2-2abcosc$

Differentiating w.r.t t

$2c\frac{dc}{dt}=0+0-2ab(-sinc)\frac{dc}{dt}$

Now, substituting the values in the above equation

$\frac{dc}{dt}=\frac{2absinc\frac{dc}{dt} }{2c}=\frac{2(1)(2)(\sqrt{3})3}{(2)(\sqrt{3})(2) }$

$\frac{dc}{dt} =3$ cm/sec

The third side c is increasing at $3$ cm/sec when the angle between the given sides is $60$° and is increasing at the rate of $3$° per second.

Answer 2

Given

A = 1cm and B = 2cm

and ∠C is 60°

To find

$\frac{dc}{dt}$

Solution

Using Cosine rule

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos C

$c^{2}$ = 1 + 4 - 4(1/4)  = 3

∴ c = √3 an when ∠ c = 60°

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2abCosc

Differentiating with respect to t

2c dc/dt = 0 + 0 - 2an(-Sin C) dc/dt

dc/dt = $\frac{2ab Sinc dc}{2c dt}$

$\frac{dc}{dt}$ = 3 cm/sec

Hence, the third side c is increasing with 3 cm/sec