Question

Can someone help me on this question and explain the process?

Answer 1

Let's call the entire angle between the 10 and 5 sides to be x

Applying to cosine rule:

`9 {}^(2) = 10 {}^(2) + 5 {}^(2) - 2(10)(5)cos(x) \\ 81 = 100 + 25 - 100cos(x) \\ 100cos(x) = 125 - 81 \\ cos(x) = (44)/(100) = 0.44 \\ x = arccos(0.44)\approx63.9`

We can conclude that the angle inside the small triangle is half of x, which i will denote by y

`y = (x)/(2) = (63.9)/(2) = 31.95`

Similarly to find the angle which i will denote by z at the right lower corner of the small or large triangle, apply the cosine law again:

`10{}^(2) = 5 {}^(2) + 9 {}^(2) - 2(9)(5)cos(z) \\ 100 = 25 + 81 - 90cos(z) \\ 90cos(z) = 106 - 100\\ cos(z) = (6)/(90) = 0.06666 \\ z = arccos(0.06666)\approx86.2`

The last angle in our small triangle which i will denote by w, will be 180 minus z and y

`w = 180 - (y + z) \\ w = 180 - (31.95 + 86.2) \\ w = 61.85`

Let t be our missing side,

applying the law of sines:

`(5)/(sin(w)) = (t)/(sin(y)) \\ t = (5sin(y))/(sin(w)) = (5sin(31.95))/(sin(86.2))`

`t \approx2.652`