Question

in a triangle with sides a=60 b=40 and c=38 the largest angle of measure is the angle opposite to the side?
Need solution step by step:

Answer 1

Integrating the secant requires a bit of manipulation.

Multiply secx by secx+tanxsecx+tanx, which is really the same as multiplying by 1. Thus, we have

∫(secx(secx+tanx)secx+tanx)dx

∫sec2x+secxtanxsecx+tanxdx

Now, make the following substitution:

u=secx+tanx

du=(secxtanx+sec2x)dx=(sec2x+secxtanx)dx

We see that du appears in the numerator of the integral, so we may apply the substitution:

∫duu=ln|u|+C

Rewrite in terms of x to get

∫secxdx=ln|secx+tanx|+C

This is an integral worth memorizing.

Answer 2

The side opposite to angle A is a = 60.

Step-by-step explanation:

According to the given information, it is given that, the sides of a triangle are a = 60, b = 40 and c = 38. Now, we need to find the side of the triangle which is just opposite to the largest angle of the triangle.

Now, we know that, when the lengths of two sides and the measurement of the included angle are known that is SAS or the lengths of all three sides that is SSS, the remaining pieces of an oblique triangle can be found using the Law of Cosines. Because we are unable to establish a solvable proportion, the Law of Sines cannot be applied in either of these situations.

Let A, B and C be the angles of the triangle.

Now, the law of cosines for the three sides of the triangle is:

a² = b²+c²-2bc(cosA) ...(1)

b² = a²+c²-2ac(cosB) ...(2)

c² = b²+a²-2ab(cosC) ...(3)

Now, equation (1) gives,

a² = b²+c²-2bc(cosA)

Or, $\frac{a^{2}-b^{2} -c^{2} }{-2bc} =cosA$

Or, $\frac{-a^{2}+b^{2} +c^{2} }{2bc} =cosA$

Or, $\frac{-60^{2}+40^{2} +38^{2} }{2*40*38} =cosA$

Or, -0.18289473684 = cos A

Or, A = 1.75473 rad = 100°32'18"

Similarly, equation (2) gives,

b² = a²+c²-2ac(cosB)

Or, $\frac{b^{2}-a^{2} -c^{2} }{-2ac} =cosB$

Or, $\frac{-b^{2}+a^{2} +c^{2} }{2ac} =cosB$

Or, $\frac{-40^{2}+60^{2} +38^{2} }{2*60*38} =cosB$

Or, B = 0.71474 rad = 40°57'6"

Similarly, equation (3) gives,

c² = b²+a²-2ab(cosC)

Or, $\frac{c^{2}-b^{2} -a^{2} }{-2ba} =cosC$

Or, $\frac{-38^{2}+40^{2} +60^{2} }{2*40*60} =cosC$

or, C = 0.67213 rad = 38°30'36"

Thus, the angle A is the largest angle.

Hence, the side opposite to angle A is a = 60.

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