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Question: 3. In the triangle ABCABC, B=90\angle B = 90^\circ, C=26.35\angle C = 26.35^\circ and \(b…

  1. In the triangle ABCABC, B=90\angle B = 90^\circ, C=26.35\angle C = 26.35^\circ and b=13.4b = 13.4 cm. Calculate the length of side cc of the triangle.

  2. In the triangle ABCABC, C=90\angle C = 90^\circ, A=69.3\angle A = 69.3^\circ and a=3.4a = 3.4 cm. Calculate the length of side cc of the triangle.

  3. An equilateral triangle has a vertical height of 20 cm. Calculate the length of the equal sides.

  4. Calculate the length of the equal sides of an isosceles triangle whose altitude (vertical height) is 15 cm and whose equal angles are 48.648.6^\circ.

Solution

Problem 3 In triangle ABC ABC , we are given: - B=90 \angle B = 90^\circ - C=26.35 \angle C = 26.35^\circ - b=13.4cm b = 13.4 \, \text{cm} We need to find the length of side c c . First, let’s find A \angle A : A=180BC \angle A = 180^\circ - \angle B - \angle C A=1809026.35=63.65 \angle A = 180^\circ - 90^\circ - 26.35^\circ = 63.65^\circ Using the sine function in right-angled triangle ABC ABC : csinA=bsinC \frac{c}{\sin \angle A} = \frac{b}{\sin \angle C} c=bsinAsinC c = \frac{b \cdot \sin \angle A}{\sin \angle C} c=13.4sin63.65sin26.35 c = \frac{13.4 \cdot \sin 63.65^\circ}{\sin 26.35^\circ} Calculating the values: sin63.650.894 \sin 63.65^\circ \approx 0.894 sin26.350.446 \sin 26.35^\circ \approx 0.446 c=13.40.8940.44626.9cm c = \frac{13.4 \cdot 0.894}{0.446} \approx 26.9 \, \text{cm} Answer: c26.9cm c \approx 26.9 \, \text{cm} --- Problem 4 In triangle ABC ABC , we are given: - C=90 \angle C = 90^\circ - A=69.3 \angle A = 69.3^\circ - a=3.4cm a = 3.4 \, \text{cm} We need to find the length of side c c . First, find B \angle B : B=180AC \angle B = 180^\circ - \angle A - \angle C B=18069.390=20.7 \angle B = 180^\circ - 69.3^\circ - 90^\circ = 20.7^\circ Using the sine function: csinA=asinC \frac{c}{\sin \angle A} = \frac{a}{\sin \angle C} c=asinAsinC c = \frac{a \cdot \sin \angle A}{\sin \angle C} c=3.4sin69.3sin90 c = \frac{3.4 \cdot \sin 69.3^\circ}{\sin 90^\circ} Calculating the values: sin69.30.939 \sin 69.3^\circ \approx 0.939 sin90=1 \sin 90^\circ = 1 c=3.40.93913.2cm c = \frac{3.4 \cdot 0.939}{1} \approx 3.2 \, \text{cm} Answer: c3.2cm c \approx 3.2 \, \text{cm} --- Problem 5 In an equilateral triangle, the vertical height (h h ) is given as 20cm 20 \, \text{cm} . We need to find the length of the equal sides. In an equilateral triangle: h=32side h = \frac{\sqrt{3}}{2} \cdot \text{side} Let s s be the length of the side: 20=32s 20 = \frac{\sqrt{3}}{2} \cdot s s=20×23 s = \frac{20 \times 2}{\sqrt{3}} s=403=403323.1cm s = \frac{40}{\sqrt{3}} = \frac{40 \sqrt{3}}{3} \approx 23.1 \, \text{cm} Answer: The equal sides are approximately 23.1cm 23.1 \, \text{cm} long. --- Problem 6 In an isosceles triangle, the altitude (h h ) is 15cm 15 \, \text{cm} , and the equal angles are 48.6 48.6^\circ . We need to find the length of the equal sides. Let s s be the length of the equal sides. The altitude divides the isosceles triangle into two congruent right-angled triangles, each with: - An angle of 48.6 48.6^\circ - Opposite side (altitude) = 15cm 15 \, \text{cm} Using the sine function: sin48.6=hs \sin 48.6^\circ = \frac{h}{s} s=hsin48.6 s = \frac{h}{\sin 48.6^\circ} s=15sin48.6 s = \frac{15}{\sin 48.6^\circ} Calculating the value: sin48.60.750 \sin 48.6^\circ \approx 0.750 s=150.750=20cm s = \frac{15}{0.750} = 20 \, \text{cm} Answer: The equal sides are 20cm 20 \, \text{cm} long.

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