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part 1: let h be the height of \\( \\triangle jkl \\). find the length …

Question

part 1: let h be the height of \\( \triangle jkl \\).
find the length of the corresponding base.
\\( \square m \\)
part 2: the measure of angle j is \\( 45.93 ^ { \circ } \\).
fill in the blank to complete the sine ratio.
\\( \sin ( 45.93 ^ { \circ } ) = \frac { h } { \square } \\)
part 3: fill in the blanks in the expression for the area of \\( \triangle jkl \\).
area of \\( \triangle jkl = \frac { 1 } { 2 } \cdot \square \cdot \square \cdot \sin ( 45.93 ^ { \circ } ) m ^ { 2 } \\)
part 4: write the area of \\( \triangle a b c \\) in terms of b, c, and a.
area of \\( \triangle a b c = \square \\)

Explanation:

Part 1:

Step1: Identify the base

The base is the side perpendicular to the height. From the figure, the base is \(7\) m.

Part 2:

Step1: Use the sine ratio formula

In a right - triangle formed by the height \(h\), the side \(15\) m (hypotenuse) and the base of the right - triangle. The sine of an angle in a right - triangle is given by \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\theta = 45.93^{\circ}\), the opposite side is \(h\) and the hypotenuse is \(15\) m. So \(\sin(45.93^{\circ})=\frac{h}{15}\)

Part 3:

Step1: Recall the area formula of a triangle

The area of a triangle \(A=\frac{1}{2}\times\text{base}\times\text{height}\). We know \(h = 15\sin(45.93^{\circ})\) (from part 2) and the base is \(7\). Substituting into the formula \(A=\frac{1}{2}\times7\times15\times\sin(45.93^{\circ})\)

Part 4:

Step1: Let the height \(h\) with respect to side \(b\)

If we consider the side \(b\) as the base. Let the height \(h\) with respect to side \(b\). If the angle at \(A\) is \(A\) and the side adjacent to angle \(A\) (in the right - triangle formed by height \(h\) and side \(c\)) is \(c\). Then \(h = c\sin A\).

Step2: Use the area formula \(A=\frac{1}{2}\times\text{base}\times\text{height}\)

Substituting the base \(b\) and height \(h = c\sin A\) into the formula \(A=\frac{1}{2}bc\sin A\)

Answer:

Part 1: \(7\)
Part 2: \(15\)
Part 3: \(7\), \(15\)
Part 4: \(\frac{1}{2}bc\sin A\)