Given data, cos(A - B) = cosAcosB + sinAsinB
<span>let, A=60' and B=30' ( here the ' sign bears degree) </span>
<span>L.H.S = cos(A - B) </span>
<span>=cos (60'-30) ( using value of A and B ) </span>
<span>=cos30' </span>
<span>= sqrt3/2 </span>
<span>R.H.S= cosAcosB + sinAsinB </span>
<span>=cos60' cos30' + sin60' sin30' </span>
<span>= 1/2* sqrt3/2+ sqrt3/2*1/2 </span>
<span>= sqrt3/4 + sqrt3/4 </span>
<span>=2 sqrt3/4 </span>
<span>= sqrt3/2 </span>
<span>so L.H.S =R.H.S or cos(A - B) = cosAcosB + sinAsinB</span>
45+55=100 180-100=80 that makes y = 80, and vertical angles makes x=80
Answer:
option C. Angle BTZ Is-congruent-to Angle BUZ
Step-by-step explanation:
Point Z is equidistant from the vertices of triangle T U V
So, ZT = ZU = ZV
When ZT = ZU ∴ ΔZTU is an isosceles triangle ⇒ ∠TUZ=∠UTZ
When ZT = ZV ∴ ΔZTV is an isosceles triangle ⇒ ∠ZTV=∠ZVT
When ZU = ZV ∴ ΔZUV is an isosceles triangle ⇒ ∠ZUV=∠ZVU
From the figure ∠BTZ is the same as ∠UTZ
And ∠BUZ is the same as ∠TUZ
So, the statement that must be true is option C
C.Angle BTZ Is-congruent-to Angle BUZ
Answer:
£495 million
Step-by-step explanation:
To find out the total cost of the land, we need to first calculate the area of the land.
Step 1: Find area of right angled triangle ADC
AD = 5 km,
DC = 12 km
Area of the right triangle = ½*a*b
a = 5km
b = 12km
Area = ½*5*12
= 5*6
Area of ADC = 30 km²
Step 2: Find the area of triangle ABC
First, let's find the length of AC using Pythagorean theorem
AC² = AD² + DC²
AC² = 5² + 12² = 25 + 144
AC = √169
AC = 13km
Area of ∆ABC = ½*AB*AC*sin(30°)
= ½*6*13*0.5
= 3*13*0.5
Area of ∆ABC = 19.5 km²
Total area of the land = area of ∆ADC + ∆ABC = 30 + 19.5 = 49.5 km²
Step 3: calculate how much the land costs
If the land costs £10 million per km²,
Cost of 49.5 km² = 49.5 × 10 = £495 million
