Answer:
x=75
Step-by-step explanation:
We can use proportions to solve this problem. Put the side of the small triangle over the side of the large triangle.
42 50
------ = ----------
42+63 50+x
42 50
------ = ----------
105 50+x
Using cross products
42(50+x) = 50*(105)
2100+42x = 5250
Subtract 2100 from each side
2100-2100+42x = 5250-2100
42x =3150
Divide by 42
42x/42 = 3150/42
x = 75
Answer:
£13,819 to the nearest pound is still £13,819.
Answer:
now just half the first triangals answers
Step-by-step explanation:
x =21
y=24
<u>TH</u>=24 <u>PQ</u>=12
<u>RH</u>=24 <u>NQ</u>=12
<u>RT</u>=21 <u>NP</u>=10.5
>T=80 >P=80
>H=20 >Q=20
>R=80 >N=80
Answer:
x = 14
Step-by-step explanation:
Assume your diagram is like the one below.
The intersecting secant angles theorem states, "When two secants intersect outside a circle, the measure of the angle formed is one-half the difference between the far and the near arcs."
For your diagram, that means
![\begin{array}{rcl}m\angle L &=&\dfrac{1}{2} \left(m \widehat {JM} - m\widehat {PQ}\right)\\\\(3x + 13)^{\circ}& = &\dfrac{1}{2} \left[(8x + 48)^{\circ} - (5x - 20)^{\circ}\right]\\\\3x + 13& = &\dfrac{1}{2}(8x + 48 - 5x + 20)\\\\3x + 13& = &\dfrac{1}{2}(3x + 68)\\\\6x + 26 & = & 3x + 68\\6x & = & 3x + 42\\3x & = & 42\\x & = & \mathbf{14}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7Dm%5Cangle%20L%20%26%3D%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%28m%20%5Cwidehat%20%7BJM%7D%20-%20m%5Cwidehat%20%7BPQ%7D%5Cright%29%5C%5C%5C%5C%283x%20%2B%2013%29%5E%7B%5Ccirc%7D%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%5B%288x%20%2B%2048%29%5E%7B%5Ccirc%7D%20-%20%285x%20-%2020%29%5E%7B%5Ccirc%7D%5Cright%5D%5C%5C%5C%5C3x%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%288x%20%2B%2048%20-%205x%20%2B%2020%29%5C%5C%5C%5C3x%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%283x%20%2B%2068%29%5C%5C%5C%5C6x%20%2B%2026%20%26%20%3D%20%26%203x%20%2B%2068%5C%5C6x%20%26%20%3D%20%26%203x%20%2B%2042%5C%5C3x%20%26%20%3D%20%26%2042%5C%5Cx%20%26%20%3D%20%26%20%5Cmathbf%7B14%7D%5C%5C%5Cend%7Barray%7D)
Check:
![\begin{array}{rcl}(3\times14 + 13) & = &\dfrac{1}{2} \left[(8\times14 + 48)^{\circ} - (5\times14 - 20)^{\circ}\right]\\\\42 + 13& = &\dfrac{1}{2}(112 + 48 - 70 + 20)\\\\55& = &\dfrac{1}{2}(110)\\\\55 & = & 55\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%283%5Ctimes14%20%2B%2013%29%20%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%20%5Cleft%5B%288%5Ctimes14%20%2B%2048%29%5E%7B%5Ccirc%7D%20-%20%285%5Ctimes14%20-%2020%29%5E%7B%5Ccirc%7D%5Cright%5D%5C%5C%5C%5C42%20%2B%2013%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%28112%20%2B%2048%20-%2070%20%2B%2020%29%5C%5C%5C%5C55%26%20%3D%20%26%5Cdfrac%7B1%7D%7B2%7D%28110%29%5C%5C%5C%5C55%20%26%20%3D%20%26%2055%5C%5C%5Cend%7Barray%7D)
It checks.
