So, what we will do is, multiply both top and bottom, by the conjugate of the denominator, so, the denominator is 7 + 10i, so its conjugate is 7 - 10i then, let's do so.
![\bf \cfrac{-3-7i}{7+10i}\cdot \cfrac{7-10i}{7-10i}\implies \cfrac{(-3-7i)(7-10i)}{(7+10i)(7-10i)}\implies \cfrac{(-3-7i)(7-10i)}{7^2-(10i)^2} \\\\\\ \cfrac{-21+30i-49i+70i^2}{7^2-(10^2i^2)}\implies \cfrac{-21+30i-49i+70\left( \boxed{-1} \right)}{7^2-\left[10^2\left( \boxed{-1} \right)\right]} \\\\\\ \cfrac{-21+30i-49i-70}{49-(-100)}\implies \cfrac{-91-19i}{149}\implies -\cfrac{91}{149}-\cfrac{19i}{149}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B-3-7i%7D%7B7%2B10i%7D%5Ccdot%20%5Ccfrac%7B7-10i%7D%7B7-10i%7D%5Cimplies%20%5Ccfrac%7B%28-3-7i%29%287-10i%29%7D%7B%287%2B10i%29%287-10i%29%7D%5Cimplies%20%5Ccfrac%7B%28-3-7i%29%287-10i%29%7D%7B7%5E2-%2810i%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B-21%2B30i-49i%2B70i%5E2%7D%7B7%5E2-%2810%5E2i%5E2%29%7D%5Cimplies%20%5Ccfrac%7B-21%2B30i-49i%2B70%5Cleft%28%20%5Cboxed%7B-1%7D%20%5Cright%29%7D%7B7%5E2-%5Cleft%5B10%5E2%5Cleft%28%20%5Cboxed%7B-1%7D%20%5Cright%29%5Cright%5D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B-21%2B30i-49i-70%7D%7B49-%28-100%29%7D%5Cimplies%20%5Ccfrac%7B-91-19i%7D%7B149%7D%5Cimplies%20-%5Ccfrac%7B91%7D%7B149%7D-%5Ccfrac%7B19i%7D%7B149%7D)
Fatima wants to mail three parcels to three village school she finds that the postal charges are rupees 20 rupees 28 and 36 resp
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Answer:
Step-by-step explanation:
Given that the lengths of the sides of a triangle are consecutive integers
Since side cannot be negative we can assume that the sides are
a, a+1 and a+2 where a>0
The largest angle is opposite side a+2 and smallest is angle opposite a
Using sine formula and the given information that the largest angle is twice the smallest angle
we get
Cross multiply to get
Since sin 2A = 2sin A cos A
we get