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Geometry 1

Find the measures of the following to the nearest tenth of a degree.  
 
¯HT   
∠T 
  
 
 Solution:
Given Data:
WH=15cm
WT=8cm
¯HT=?
∠T=?
Sol: A
As we know here in the following figure (HWT) is a right angle triangle, where WT is base & WH is perpendicular and HT is hypotenuses. So, according to Pythagoras theorem, 
(Hypotenuses)^2=(Base)^2+(Perpendicular)^2
(HT)^2=(WT)^2+(WH)^2
As here Base and perpendicular are known and to find Hypotenuses, Let, WT=a, WH=band HT=c. Hence, 
a^2+b^2=c^2
a=8
b=15
c=?
To find c (HT) put values of a and b in the equation,
a^2+b^2=c^2
8^2+15^2=c^2
c^2=8^2+15^2
c^2=64+225
c^2=289
To find ¯HT take the square root of  (∴c=HT)
(HT)^2=√289
¯HT=√289
¯HT=17Ans
Sol: B
As we know tan⁡ θ=Perpendicular/Base and∠T=〖tan〗^(-1)⁡〖Perpendicular/Base〗. So, to find ∠Tput given values in the following equation 
∠T=〖tan〗^(-1)⁡〖WH/WT〗
∠T=〖tan〗^(-1)⁡〖15/8〗
∠T=〖tan〗^(-1)⁡(1.875)
∠T=61.927^∘≈61.93^∘ Ans

What is the area of the given triangle? 
Solution:
Given Data:
BC=13cm
CA=7cm
θ=38^∘
Areaoftringle=?
Sol: 
As we know to find area of triangle, we apply following formula, 
A=1/2×a×b×sin⁡ θ
Put value of a (BC) and b (CA) and   in equation to find area of triangle,
A=1/2×a×b×sin⁡ θ
A=1/2×13cm×7cm×sin⁡3 8^∘
A=1/2×91cm^2×sin⁡3 8^∘
A=1/2×91×"0.61566148c" "m" ^2
A="56.025" /2 "c" "m" ^2
A="28.01c" "m" ^2 Ans
If X and Y are complementary angles, sin⁡X=15/17 , and cos⁡X=8/17, find each of the following: 
tan X = 
sin Y = 
cos Y = 
tan Y = 
Solution:
Given Data:
sin⁡X=15/17
cos⁡X=8/17
To find:
tan⁡X=?
sin⁡Y=?
cos⁡Y=?
tan⁡Y=?
Sol:
According to the given data, it comes to know that 
Base=15,Perpendicular=8&Hypotenuses=17
So, by drawing the right angle triangle,
 
tan X=?
As we know that tan⁡X=sin⁡X/cos⁡X, so putting values of sin & cos we can find tan X
tan⁡X=sin⁡X/cos⁡X 
tan⁡X=(15/17)/(8/17)
∴tan⁡X=15/8 Ans
sin Y=?
As we knew that X=〖tan〗^(-1)⁡〖15/8〗=61.93^∘since x and y are complementary angles. So, 
X+Y=90
61.93+Y=90
Y=90-61.93
Y=28.07
So, for sin Y is, 
sin⁡Y=sin⁡( 28.07)
sin⁡Y=8/17 Ans

cos Y=?
Similarly, we can calculate cos Y by putting a value of Y in it, 
cos⁡Y=cos⁡( 28.07)
cos⁡Y=15/17 Ans


tan Y=?
tan⁡Y=sin⁡Y/cos⁡Y 
tan⁡Y=(8/17)/(15/17)
tan⁡Y=8/15=0.5333Ans
 

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