+58 Find AC in the diagram to the nearest integer.
\(In $\triangle PQR,$ we have $\angle P = 90^\circ,$ $PQ=3,$ and $QR=\sqrt {58}.$ Find $\tan R.$ \)
\(Find the value of $\sin 40^\circ \cdot \cos 50^\circ + \cos 40^\circ \cdot \sin 50^\circ.$\)(Says: "Find the value of...")
In the diagram below, we know \(\tan \theta = \frac{3}{4}. \) Find the area of the triangle.
+9466 Find AC in the diagram to the nearest integer.
sin( angle ) = opposite / hypotenuse
sin( 70° ) = 31 / AC
Multiply both sides of the equation by AC
AC sin( 70° ) = 31
Divide both sides of the equation by sin( 70° )
AC = 31 / sin( 70° )
Plug 31 / sin( 70° ) into a calculator.
AC ≈ 33
+9466 In \(\triangle PQR,\) we have \(\angle P = 90^\circ,\) \(PQ=3,\) and \(QR=\sqrt {58}.\) Find \(\tan R.\)
By the Pythagorean theorem,
32 + ( PR )2 = √[ 58 ]2
9 + ( PR )2 = 58
( PR )2 = 49
PR = 7
tan( angle ) = opposite / adjacent
tan( R ) = 3 / PR
tan( R ) = 3 / 7
+9466 Find the value of \(\sin40°\cdot\cos50°+\cos40°\cdot\sin50°\) .
Remember the angle addition formula for sine: \(\sin(\alpha+\beta)\ =\ \sin\alpha\cos\beta+\cos\alpha\sin\beta\)
So...
\(\sin40°\cdot\cos50°+\cos40°\cdot\sin50°\ =\ \sin(40°+50°)\ =\ \sin(90°)\ =\ 1\)
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+9466 Part of this last problem is just like the second problem.......
In the diagram below, we know \(\tan\theta=\frac34\) . Find the area of the triangle.
Let the base of the triangle be the side with length 60. Draw a height to that base and call it h .
By the Pythagorean Identity,
| 1+ cot2θ = csc2θ |
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| 1 + ( 4/3 )2 = 1 / sin2θ |
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| 25 / 9 = 1 / sin2θ |
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| 9 / 25 = sin2θ |
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| sin θ = 3 / 5 |
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| sin( angle ) = opposite / hypotenuse | |
| sin θ = h / 40 |
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| 3 / 5 = h / 40 |
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| 24 = h |
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| area of triangle = (1/2)(base)(height) | |
area of triangle = (1/2)(60)(24) |
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| area of triangle = 720 |
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