On a baseball field, the pitcher’s mound is 60.5 feet from home plate. During practice, a batter hits a ball 195 feet at an angl
e of 32° to the right of the pitcher’s mound. An outfielder catches the ball and throws it to the pitcher. Approximately how far does the outfielder throw the ball?
1 answer:
The problem is modelled in the diagram below
We will use the cosine rule to find x
x² = 60.5² + 195² - [2×60.5×195×cos(32°)]
x² = 3660.25 + 38025 - [23595 × cos(32)]
x² = 41685.25 - 19683.50019
x² = 22001.74981
x = √22001.74981
x = 148.33 feet (rounded to 2 decimal place)
Note: the cosine rule is shown in the second diagram below
We will use the cosine rule to find x
x² = 60.5² + 195² - [2×60.5×195×cos(32°)]
x² = 3660.25 + 38025 - [23595 × cos(32)]
x² = 41685.25 - 19683.50019
x² = 22001.74981
x = √22001.74981
x = 148.33 feet (rounded to 2 decimal place)
Note: the cosine rule is shown in the second diagram below
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Answer:
c = 6.1
Step-by-step explanation:
Given that: B = 18° , C = 152°, and b = 4.
The missing side, c, can be determined by applying the sine rule which states that:
=
=
c =
=
= 6.0773
c = 6.1
Therefore, the value of the missing side, c, is 6.1
Answer:
By AA
ΔWXY ~ΔWVZ
Step-by-step explanation:
Here WXY is an isosceles triangle with legs WX & WY
So WX = WY
Hence ∠X = ∠Y
So ∠2= ∠3.
Now by angle sum property
∠1 + ∠2+∠3 = 180°
∠1+∠2+∠2=180°
2∠2 = 180° - ∠1 .......(1)
In triangle WVZ
WV = WZ
So ∠V = ∠Z
∠4 = ∠5
Once again by angle sum property
∠1 + ∠4 + ∠5=180°
∠1 + ∠4 + ∠4 = 180°
2∠4 = 180° - ∠1 ...(2)
From (1) & (2)
2∠2 = 2∠4
∠2=∠4
Now ∠W is common to both triangles
Hence by AA
ΔWXY ~ΔWVZ
